Differential Equations | CBSE Class 12 Maths Notes

Differential Equations in Mathematics covers equations involving derivatives, order and degree, general and particular solutions, verification of solutions, separation of variables, homogeneous equations, linear equations, integrating factors, tangent conditions and continuous growth.

What is a differential equation?

A differential equation is an equation involving derivatives of a dependent variable with respect to one or more independent variables. It connects an unknown function with the rate at which that function changes. Solving it means finding a function that satisfies this relationship.

Throughout the usual form of these equations, $x$ denotes the independent variable and $y$ the dependent variable. The symbol $f$ denotes a function, so $y=f(x)$ means that the value of $y$ depends on $x$.

Definition: An ordinary differential equation involves derivatives with respect to just one independent variable. A partial differential equation involves derivatives with respect to more than one independent variable. The methods below concern ordinary differential equations.

How are the derivatives written?

The notation $y'$ means the first derivative, $y''$ the second derivative and $y'''$ the third derivative, all with respect to $x$. In full notation these are $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$ and $\frac{d^3y}{dx^3}$, respectively.

For example, $x+y=7$ contains no derivative, whereas $\frac{dy}{dx}+y=0$ does. This is the essential distinction between an ordinary algebraic relation involving variables and a differential equation.

Differentiation starts with a function and finds its derivative. Solving a differential equation works in the reverse direction: the derivative relationship is given, and a function satisfying it is sought. Integration is therefore central to the methods used here.

How do order and degree differ?

The order of a differential equation is the order of the highest derivative appearing in it. Read the derivative itself before looking at its power. Squaring a first derivative does not turn it into a second derivative.

The degree is defined when the equation is polynomial in its derivatives. It is the highest power of the highest order derivative in that polynomial equation. A trigonometric or exponential function of a derivative can prevent degree from being defined.

Note: The polynomial condition concerns derivatives. A term such as $\sin y$ does not itself prevent degree from being defined, because $y$ is the dependent variable rather than one of its derivatives.

How should a mixed-power equation be classified?

Worked example 1. Find the order and degree of $(y''')^2+(y'')^3+(y')^4+y^5=0$.

Answer:

1. Identify all derivatives: $y'$, $y''$ and $y'''$. The highest derivative is $\frac{d^3y}{dx^3}$.
2. The order is therefore $3$, regardless of the other exponents.
3. The equation is polynomial in its derivatives. The highest derivative occurs as $(y''')^2$, so the degree is $2$.
4. Check the tempting alternatives: the power $4$ belongs to the first derivative, while $5$ belongs to the dependent variable. Neither determines the degree.

What are general and particular solutions?

A solution is a function that satisfies the differential equation when the function and its derivatives are substituted into it. Verification is an identity check: the two sides must agree throughout the domain being considered.

A general solution, also called a primitive, contains arbitrary constants. In the general solutions considered here, the number of independent arbitrary constants equals the order of the differential equation. A particular solution fixes those constants and contains no arbitrary constants.

How can a family satisfy one equation?

Let $a$ and $b$ denote arbitrary real constants. The family $y=a\cos x+b\sin x$ is a general solution of $y''+y=0$. Giving particular values to $a$ and $b$ selects a member of that family.

Worked example 2. Verify $y=a\cos x+b\sin x$ for $y''+y=0$, where $a$ and $b$ are arbitrary real constants.

Answer:

1. Differentiate once, keeping the constants fixed: $$y'=-a\sin x+b\cos x.$$
2. Differentiate again: $$y''=-a\cos x-b\sin x.$$
3. Substitute both the original function and its second derivative: $$y''+y=(-a\cos x-b\sin x)+(a\cos x+b\sin x)=0.$$
4. The terms cancel for every allowed value of $x$. Thus the family satisfies the equation and retains the two arbitrary constants appropriate to this second order equation.

How is an exponential solution verified?

Worked example 3. Verify $y=e^{-3x}$ as a solution of $y''+y'-6y=0$. Here $e$ is the base of natural logarithms.

Answer:

1. Use the chain rule: $$y'=-3e^{-3x}.$$
2. Differentiate the first derivative: $$y''=9e^{-3x}.$$
3. Substitute into the complete left side: $$y''+y'-6y=9e^{-3x}-3e^{-3x}-6e^{-3x}=0.$$
4. The right side is also $0$. The proposed function is therefore a solution, with no arbitrary constant left to determine.

How does separation of variables work?

A first order, first degree equation may be written $\frac{dy}{dx}=F(x,y)$, where $F$ denotes a function of the two variables. It is separable when the right side is a product of a function of $x$ and a function of $y$.

Write the factors as $g(x)$ and $h(y)$, where $g$ depends only on $x$ and $h$ only on $y$. The purpose of separation is to place each variable beside its own differential before integration.

Result: the separated integral form

1. Start with the product form: $$\frac{dy}{dx}=g(x)h(y).$$
2. On an interval where $h(y)\ne0$, divide by this factor and separate: $$\frac{1}{h(y)}\,dy=g(x)\,dx.$$
3. Integrate each side with respect to the variable shown: $$\int\frac{1}{h(y)}\,dy=\int g(x)\,dx.$$
4. Let $H$ be an antiderivative of $1/h$, $G$ an antiderivative of $g$, and $C$ an arbitrary constant. The solution is $$H(y)=G(x)+C.$$

One arbitrary constant suffices because the difference of the two constants introduced by integration can be replaced by a single constant. A solution can remain implicit; isolating the dependent variable is not required in every problem.

How is an implicit solution obtained?

Worked example 4. Solve $\frac{dy}{dx}=\frac{x+1}{2-y}$, with $y\ne2$.

Answer:

1. Separate the variables: $$(2-y)\,dy=(x+1)\,dx.$$
2. Integrate both sides: $$2y-\frac{y^2}{2}=\frac{x^2}{2}+x+C_1,$$ where $C_1$ is an arbitrary integration constant.
3. Multiply through by two, rearrange, and write $C=2C_1$: $$x^2+y^2+2x-4y+C=0.$$
4. Differentiate the answer to check it: $$2x+2yy'+2-4y'=0.$$
5. Collect the derivative terms: $$(2-y)y'=x+1.$$ Dividing by $2-y$ recovers the given equation on its stated domain.

How are particular solutions found after separation?

An initial condition gives the dependent variable at a specified value of the independent variable. First integrate to obtain a family of solutions. Then substitute the given values into that family to calculate the arbitrary constant.

Keep the integration and the use of the condition as separate steps. Substituting the initial values into the differential equation gives information at one point; it does not replace finding a function that satisfies the equation.

What happens when inverse trigonometric functions appear?

Worked example 5. Find the general solution of $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$.

Answer:

1. Since $1+y^2>0$ for real $y$, separate without dividing by zero: $$\frac{dy}{1+y^2}=\frac{dx}{1+x^2}.$$
2. Integrate both sides: $$\tan^{-1}y=\tan^{-1}x+C.$$ The notation $\tan^{-1}$ denotes the inverse tangent function.
3. Check by implicit differentiation: $$\frac{y'}{1+y^2}=\frac{1}{1+x^2}.$$
4. Multiply by $1+y^2$ to recover the equation. The arbitrary constant remains because no initial condition was supplied.

How is the constant determined?

Worked example 6. Solve $\frac{dy}{dx}=-4xy^2$, given $y=1$ when $x=0$.

Answer:

1. For the nonzero solution through the given point, separate: $$\frac{dy}{y^2}=-4x\,dx.$$
2. Integrate: $$-\frac1y=-2x^2+C.$$
3. Insert the condition: $$-1=0+C,\qquad C=-1.$$
4. Substitute the constant and solve for the dependent variable: $$y=\frac{1}{2x^2+1}.$$
5. Recompute its derivative: $$y'=-\frac{4x}{(2x^2+1)^2}=-4xy^2.$$ Also, $y(0)=1$, so both requirements hold.

Note: Dividing by a dependent-variable factor can exclude a solution. Here $y=0$ also satisfies the original differential equation, but cannot satisfy the supplied initial condition.

How do tangent conditions produce differential equations?

The slope of the tangent to a curve at a point $(x,y)$ is $\frac{dy}{dx}$. A statement giving that slope therefore supplies a differential equation. A point through which the curve passes supplies the condition needed to determine a particular member of the solution family.

Translate the two pieces of information separately. The slope relationship applies at points along the curve. The supplied point is used after integration to identify the required curve. Its coordinates must satisfy the final answer.

How is a curve recovered from its slope?

Worked example 7. Find the curve through $(-2,3)$ whose tangent slope at $(x,y)$ is $\frac{2x}{y^2}$.

Answer:

1. Translate the slope statement: $$\frac{dy}{dx}=\frac{2x}{y^2},\qquad y\ne0.$$
2. Separate the variables: $$y^2\,dy=2x\,dx.$$
3. Integrate: $$\frac{y^3}{3}=x^2+C.$$
4. Use the point: $$\frac{3^3}{3}=(-2)^2+C,\qquad 9=4+C,\qquad C=5.$$
5. Write the required curve: $$y^3=3x^2+15,\qquad y=(3x^2+15)^{1/3}.$$
6. Differentiate the implicit equation: $$3y^2y'=6x,\qquad y'=\frac{2x}{y^2}.$$ At the supplied point, both sides of $y^3=3x^2+15$ equal $27$.

The implicit form makes the derivative check short, while the explicit form gives the dependent variable directly. These are two ways of presenting the same particular curve. In either form, retain restrictions required by the original slope expression.

What makes a differential equation homogeneous?

A function $F(x,y)$ is homogeneous of degree $n$ if simultaneous scaling of both variables gives $F(\lambda x,\lambda y)=\lambda^nF(x,y)$. Here $n$ denotes the degree of homogeneity and $\lambda$ is a nonzero scale factor.

This degree describes a function under scaling. It is different from the degree of a differential equation, which describes the power of the highest derivative. Keeping these two definitions separate prevents a common classification error.

Result: the ratio test for a homogeneous equation

A differential equation written $\frac{dy}{dx}=F(x,y)$ is homogeneous when $F$ is homogeneous of degree zero. On a domain where $x\ne0$, its right side can be expressed as $g(y/x)$, a function of the ratio alone.

For example, $\frac{dy}{dx}=\frac{x+2y}{x-y}$ has right side $\frac{1+2(y/x)}{1-y/x}$. The cancellation of a common scale factor shows why the ratio is useful. The denominator also requires $x\ne y$.

The expression $\sin x+\cos y$ does not satisfy the same scaling condition. Recognise homogeneity by testing the function rather than by assuming that every equation containing both variables is homogeneous.

How does the homogeneous substitution make variables separable?

For an equation whose right side depends on $y/x$, introduce a new dependent variable $v$ by $v=y/x$. Thus $y=vx$, where $v$ varies with $x$. The product rule is essential when differentiating this substitution.

Derivation: reduction to a separable equation

1. Start with $\frac{dy}{dx}=g(y/x)$ and substitute $y=vx$.
2. Differentiate the product: $$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$
3. Replace the ratio by $v$ and rearrange: $$v+x\frac{dv}{dx}=g(v),\qquad x\frac{dv}{dx}=g(v)-v.$$
4. Where the divisors are nonzero, separate and integrate: $$\int\frac{dv}{g(v)-v}=\int\frac{dx}{x}+C.$$
5. Replace $v$ by $y/x$ to express the answer in the original variables.

Treat the substitution variable as a function of $x$. Use the product rule rather than assuming its derivative is zero. Constant values of $v$ satisfying $g(v)=v$ can give straight-line solutions $y=vx$; check these separately in the original equation before dividing by $g(v)-v$. The last substitution is just as necessary: an answer in the auxiliary variable alone does not yet give the requested relation between the original variables.

How does this work with a trigonometric ratio?

Worked example 8. Solve $x\cos(y/x)\frac{dy}{dx}=y\cos(y/x)+x$, on a domain where $x\ne0$ and $\cos(y/x)\ne0$.

Answer:

1. Divide to obtain $$y'=\frac{y}{x}+\sec(y/x).$$ The symbol $\sec$ denotes the reciprocal cosine function. The right side depends on the ratio alone.
2. Set $y=vx$, so $y'=v+xv'$, with $v'=dv/dx$. Then $$v+xv'=v+\sec v.$$
3. Subtract $v$ and separate: $$\cos v\,dv=\frac{dx}{x}.$$
4. Integrate and return to the original variables: $$\sin v=\ln|x|+C,\qquad \sin(y/x)=\ln|x|+C.$$ Here $\ln$ denotes the natural logarithm.
5. Check implicitly: $$\cos(y/x)\frac{xy'-y}{x^2}=\frac1x.$$ Multiplication by $x^2$ gives the original equation after rearrangement.

What is a first order linear differential equation?

A first order linear differential equation in the dependent variable $y$ has the standard form $\frac{dy}{dx}+P(x)y=Q(x)$. Here $P$ is the coefficient function of the dependent variable and $Q$ is the function on the right side. Both depend only on $x$, or are constants.

Before identifying these functions, make the coefficient of the derivative equal to one. The integrating factor is a multiplier that turns the complete left side into the derivative of a product.

Result: the integrating-factor formula

Let $I(x)$ denote the integrating factor. Its formula is $I(x)=e^{\int P(x)\,dx}$. The solution is $yI(x)=\int Q(x)I(x)\,dx+C$. The following derivation explains why the same factor multiplies every term.

1. Multiply the standard equation by $I$: $$Iy'+PIy=QI.$$
2. Apply the product rule to the intended left side: $$\frac{d}{dx}(Iy)=Iy'+I'y,$$ where $I'$ means the derivative of the integrating factor.
3. Match coefficients by choosing $I'=PI$. Dividing by the nonzero factor and integrating gives $$\frac{I'}I=P,\qquad I=e^{\int P\,dx}.$$
4. The multiplied equation becomes $$\frac{d}{dx}(Iy)=QI.$$
5. Integrate once more to obtain $$Iy=\int QI\,dx+C.$$

Check the product derivative before integrating. It confirms both the sign of the coefficient and the integrating factor. The final integration introduces the arbitrary constant required for the general solution.

Note: In this method, a term involving a power of the dependent variable greater than one does not fit the stated linear form. The coefficient functions must depend on the chosen independent variable alone.

How are linear equations solved step by step?

The standard sequence is to normalise the derivative coefficient, identify the coefficient functions, find the integrating factor, recognise a product derivative and integrate. Keeping these stages visible makes sign errors easier to detect.

How is a non-unit derivative coefficient handled?

Worked example 9. Solve $x\frac{dy}{dx}+2y=x^2$, with $x\ne0$.

Answer:

1. Divide the whole equation by $x$: $$y'+\frac2x y=x.$$ Thus $P(x)=2/x$ and $Q(x)=x$.
2. Find an integrating factor: $$I=e^{\int(2/x)\,dx}=e^{2\ln|x|}=x^2.$$
3. Multiply through and recognise the product: $$x^2y'+2xy=x^3,\qquad \frac{d}{dx}(x^2y)=x^3.$$
4. Integrate and divide by $x^2$: $$x^2y=\frac{x^4}{4}+C,\qquad y=\frac{x^2}{4}+\frac{C}{x^2}.$$
5. Differentiate: $$y'=\frac{x}{2}-\frac{2C}{x^3}.$$ Substitution gives $$xy'+2y=\frac{x^2}{2}-\frac{2C}{x^2}+\frac{x^2}{2}+\frac{2C}{x^2}=x^2.$$

How is integration by parts used?

Worked example 10. Solve $\frac{dy}{dx}-y=\cos x$.

Answer:

1. Identify $P=-1$, $Q=\cos x$, and calculate $I=e^{-x}$.
2. Multiply and integrate: $$\frac{d}{dx}(ye^{-x})=e^{-x}\cos x,\qquad ye^{-x}=\int e^{-x}\cos x\,dx+C.$$
3. Let $J$ denote the integral on the right, omitting its constant temporarily. Integration by parts gives $$J=-e^{-x}\cos x-\int e^{-x}\sin x\,dx.$$
4. Apply integration by parts to the remaining integral: $$\int e^{-x}\sin x\,dx=-e^{-x}\sin x+J.$$
5. Substitute back and collect the two copies of $J$: $$2J=e^{-x}(\sin x-\cos x).$$
6. Insert the integral and multiply by $e^x$: $$y=\frac{\sin x-\cos x}{2}+Ce^x.$$
7. Recompute the derivative: $$y'=\frac{\cos x+\sin x}{2}+Ce^x.$$ Subtracting the solution gives $y'-y=\cos x$, as required.

The factor comes from the signed coefficient of the dependent variable. In the second example that coefficient is negative. Reversing its sign would prevent the multiplied left side from being the intended product derivative.

When is it better to treat the other variable as dependent?

Some equations become linear when $x$ is treated as a function of $y$. Their standard form is $\frac{dx}{dy}+P_1(y)x=Q_1(y)$, where $P_1$ and $Q_1$ denote coefficient functions depending only on $y$, or constants.

All stages then use $y$ as the independent variable. The integrating factor is $e^{\int P_1(y)\,dy}$, and the quantity multiplied by it in the solution formula is the dependent variable $x$.

How does the reversed linear form work?

Worked example 11. Solve $y\,dx-(x+2y^2)\,dy=0$, treating $x$ as a function of $y$ on an interval where $y\ne0$.

Answer:

1. Divide and arrange the linear form: $$\frac{dx}{dy}-\frac{x}{y}=2y.$$ Here $P_1=-1/y$ and $Q_1=2y$.
2. Choose the integrating factor $I=1/y$. Direct differentiation verifies $$\frac{dI}{dy}=-\frac1{y^2}=P_1I.$$
3. Multiply through: $$\frac1y\frac{dx}{dy}-\frac{x}{y^2}=2,\qquad \frac{d}{dy}\left(\frac{x}{y}\right)=2.$$
4. Integrate and rearrange: $$\frac{x}{y}=2y+C,\qquad x=2y^2+Cy.$$
5. Check using $dx/dy=4y+C$: $$y\frac{dx}{dy}-x-2y^2=y(4y+C)-(2y^2+Cy)-2y^2=0.$$

The choice of dependent variable determines which coefficient functions are allowed and which integration variable to use. Similarly, a homogeneous equation written as $dx/dy=h(x/y)$ suggests the substitution $x=vy$, where now $v$ depends on $y$.

How are initial conditions used in linear equations?

The integrating-factor method first produces a family of solutions. A condition such as a specified point then fixes its arbitrary constant. The method does not change, but the final answer must satisfy both the differential equation and that point.

How is a tangent problem solved by a linear method?

Worked example 12. Find the curve through $(0,1)$ whose tangent slope at $(x,y)$ equals the sum of the horizontal coordinate and the product of the two coordinates.

Answer:

1. Translate and rearrange: $$y'=x+xy,\qquad y'-xy=x.$$
2. Identify $P=-x$ and $Q=x$. The integrating factor is $$I=e^{\int -x\,dx}=e^{-x^2/2}.$$
3. Multiply to form a product derivative: $$\frac{d}{dx}\left(ye^{-x^2/2}\right)=xe^{-x^2/2}.$$
4. Use the derivative of the exponential to integrate: $$\frac{d}{dx}\left(e^{-x^2/2}\right)=-xe^{-x^2/2},\qquad ye^{-x^2/2}=-e^{-x^2/2}+C.$$
5. Multiply by $e^{x^2/2}$, then insert the point: $$y=-1+Ce^{x^2/2},\qquad 1=-1+C,\qquad C=2.$$
6. The required curve is $$y=-1+2e^{x^2/2}.$$ Its derivative is $y'=2xe^{x^2/2}=x+xy$, and it gives $y(0)=1$.

A useful final check has two parts. Differentiating verifies the rate relationship everywhere the formula applies. Substituting the initial point verifies the constant. Neither check by itself establishes both requirements of a particular-solution problem.

Recognising the derivative of the exponential also simplifies the integration. Its exponent has derivative $-x$, which explains the negative sign in the antiderivative used above.

How does continuous growth lead to a differential equation?

When the rate of increase of a quantity is proportional to the quantity present, a differential equation expresses that relationship. Continuous growth is handled by integrating a rate equation, with the initial quantity determining the arbitrary constant.

In the following example, $P$ denotes the principal amount in rupees, rather than the coefficient function used in the linear-equation sections. The variable $t$ denotes elapsed time in years. The stated rate is a rate of increase of the current amount.

How long does the given principal take to double?

Worked example 13. A bank principal increases continuously at $5\%$ per year. Find the time in which ₹$1000$ doubles.

Answer:

1. Translate the rate: $$\frac{dP}{dt}=\frac5{100}P=\frac{P}{20},\qquad P(0)=1000.$$
2. Since the principal is positive, separate and integrate: $$\frac{dP}{P}=\frac{dt}{20},\qquad \ln P=\frac{t}{20}+C_1.$$
3. Let $A=e^{C_1}$ be the positive multiplicative constant. Exponentiation gives $$P=Ae^{t/20}.$$
4. Use the initial amount: $$1000=Ae^0,\qquad P=1000e^{t/20}.$$
5. For doubling, set $P=2000$: $$2000=1000e^{t/20},\qquad 2=e^{t/20},\qquad t=20\ln2.$$ The answer is $20\ln2$ years.
6. Check the model by differentiation: $$\frac{dP}{dt}=50e^{t/20}=\frac{P}{20}.$$ At the computed time, $P=1000e^{\ln2}=2000$.

The initial amount and the target amount perform different roles: one fixes the solution, while the other determines the required time. Keep the time unit consistent with the period used in the stated percentage rate.

Glossary

• Differential equation — An equation involving derivatives of a dependent variable with respect to one or more independent variables.
• Ordinary differential equation — A differential equation involving derivatives with respect to just one independent variable.
• Order — The order of the highest derivative appearing in the given differential equation.
• Degree — The highest power of the highest order derivative, when the equation is polynomial in its derivatives.
• Solution — A function that satisfies a differential equation when its derivatives and values are substituted.
• General solution — A solution containing arbitrary constants, with their number corresponding to the order in the equations considered.
• Particular solution — A solution obtained by fixing the arbitrary constants of the general solution at particular values.
• Primitive — Another name for the general solution of a differential equation containing arbitrary constants.
• Separable equation — An equation in which the variables can be separated completely beside their respective differentials before integration.
• Homogeneous function — A function whose simultaneous scaling of all variables produces a fixed power of the scale factor.
• Linear differential equation — A first order equation linear in the dependent variable, with coefficients depending on the independent variable alone.
• Integrating factor — A multiplier that converts the left side of a linear differential equation into a product derivative.

Common errors and misconceptions

• Misconception: The largest exponent anywhere gives the order. Correct: Order depends on the highest derivative, not its exponent or the power of the dependent variable.
• Misconception: A sine term always makes degree undefined. Correct: A sine of the dependent variable can occur in a polynomial equation in derivatives; a sine of a derivative prevents that polynomial form.
• Misconception: Any constant appearing in an equation is arbitrary. Correct: An arbitrary constant is a free parameter; a particular solution fixes its value using supplied information.
• Misconception: Differentiating $y=vx$ gives $y'=v$. Correct: The substitution variable depends on $x$, so the product rule gives $y'=v+x\frac{dv}{dx}$.
• Misconception: The integrating factor can be read before normalising the derivative coefficient. Correct: First put the equation into standard linear form and then identify the signed coefficient of the dependent variable.
• Misconception: One multiplies just the derivative term by the integrating factor. Correct: Multiply every term, so the entire left side becomes a product derivative.
• Misconception: A factor involving the dependent variable may be cancelled without checking zero. Correct: Check excluded values in the original equation, then check whether they satisfy any given initial condition.
• Misconception: A solution satisfying the given point is automatically correct. Correct: It must also satisfy the differential equation when differentiated and substituted.

Exam-style questions with model answers

Q1. Determine the order and degree of $y''+(y')^2+2y=0$. Explain why the squared first derivative does not decide the degree. [2 marks]

1. The highest derivative present is $y''$, so the order is $2$.
2. The equation is polynomial in its derivatives and $y''$ has power $1$, so the degree is $1$. The squared term concerns a lower order derivative.

Q2. Verify that $y=e^{-3x}$ satisfies $y''+y'-6y=0$. [3 marks]

1. Differentiate the proposed function with respect to the independent variable, using the chain rule: $y'=-3e^{-3x}$.
2. Differentiate again to obtain the second derivative required by the equation: $y''=9e^{-3x}$.
3. Substitute all three expressions into the left side: $9e^{-3x}-3e^{-3x}-6e^{-3x}=0$. This equals the right side for every real value of $x$, verifying the proposed solution.

Q3. Find the particular solution of $\frac{dy}{dx}=-4xy^2$, given $y=1$ at $x=0$. [3 marks]

1. For the nonzero solution through the supplied point, separate the variables: $y^{-2}\,dy=-4x\,dx$.
2. Integrate to get $-1/y=-2x^2+C$, where $C$ is an arbitrary constant.
3. Use the initial condition: $-1=C$. Hence $y=1/(2x^2+1)$.
4. Check the answer by differentiation: $y'=-4x/(2x^2+1)^2=-4xy^2$. Substitution at the initial point also gives $y(0)=1$, so the rate equation and the condition are both satisfied.

Q4. Solve $x\frac{dy}{dx}+2y=x^2$, where $x\ne0$, using an integrating factor, and verify your solution. [5 marks]

1. Divide every term by the nonzero independent variable to obtain the standard linear form $y'+(2/x)y=x$. The coefficient function is $P=2/x$ and the right side is $Q=x$.
2. Calculate the integrating factor $I=e^{\int(2/x)\,dx}=e^{2\ln|x|}=x^2$. The absolute value keeps the logarithmic step valid on either permitted side of zero.
3. Multiply the standard equation by this factor. The left side becomes a product derivative: $\frac{d}{dx}(x^2y)=x^3$.
4. Integrate to obtain $x^2y=x^4/4+C$, then divide by $x^2$: $y=x^2/4+C/x^2$.
5. Recompute $y'=x/2-2C/x^3$. Substitution gives $xy'+2y=x^2$, since the terms containing the arbitrary constant cancel. This verifies the answer on intervals excluding zero.

Q5. Show that $x\cos(y/x)\frac{dy}{dx}=y\cos(y/x)+x$ is homogeneous and solve it where $x\ne0$ and $\cos(y/x)\ne0$. [5 marks]

1. Divide to write $y'=y/x+\sec(y/x)$. The right side is a function of the ratio $y/x$ alone, so simultaneous scaling of the two variables leaves it unchanged. The equation is homogeneous.
2. Introduce $v=y/x$, a function of $x$, so $y=vx$. The product rule gives $y'=v+x\frac{dv}{dx}$.
3. Substitute and cancel the matching terms: $v+x\frac{dv}{dx}=v+\sec v$. Thus $\cos v\,dv=dx/x$.
4. Integrate to get $\sin v=\ln|x|+C$, then restore the original variables: $\sin(y/x)=\ln|x|+C$.
5. For verification, implicit differentiation gives $\cos(y/x)(xy'-y)/x^2=1/x$. Multiplying by $x^2$ and rearranging reproduces the given differential equation, within the domain stated in the question. The constant remains arbitrary because no initial value has been supplied.

Q6. Solve $y\,dx-(x+2y^2)\,dy=0$ as a linear equation for $x$ in terms of $y$, on an interval where $y\ne0$. [4 marks]

1. Write $dx/dy-x/y=2y$. This is linear in $x$, with coefficient $-1/y$. In this orientation, every differentiation and integration uses the other coordinate as the independent variable.
2. Choose integrating factor $I=1/y$; it satisfies $dI/dy=-1/y^2=(-1/y)I$.
3. Multiply the complete equation by the factor to obtain $\frac{d}{dy}(x/y)=2$.
4. Integrate: $x/y=2y+C$, giving $x=2y^2+Cy$. Differentiating gives $dx/dy=4y+C$; substitution into $y(dx/dy)-x-2y^2$ gives zero, confirming the solution.

Q7. A principal of ₹$1000$ increases continuously at $5\%$ per year. Form the differential equation and find the exact time needed for the amount to double. [5 marks]

1. Let $P$ denote the principal in rupees and $t$ elapsed time in years. The proportional growth condition gives $dP/dt=P/20$, with initial condition $P(0)=1000$.
2. Since the amount is positive, separate and integrate: $dP/P=dt/20$, giving $\ln P=t/20+C_1$.
3. Exponentiate, writing $A=e^{C_1}$: $P=Ae^{t/20}$. The initial condition makes $A=1000$, so the amount at time $t$ is $P=1000e^{t/20}$.
4. Doubling means $P=2000$. Therefore $e^{t/20}=2$, and the required time is $t=20\ln2$ years.
5. Check by substituting this time into the amount formula: $1000e^{\ln2}=2000$. Its derivative also equals $P/20$, confirming the continuous growth relationship used throughout. The time unit is years, matching the annual rate given in the question.

Q8. Find the curve through $(1,1)$ satisfying $x\,dy=(2x^2+1)\,dx$, where $x\ne0$. [3 marks]

1. Divide by the nonzero independent variable: $dy=(2x+1/x)\,dx$.
2. Integrate term by term to obtain $y=x^2+\ln|x|+C$, where $C$ denotes an arbitrary constant.
3. Insert the supplied point: $1=1+\ln1+C$, so $C=0$. The required curve is $y=x^2+\ln|x|$, on the positive interval containing the given point.
4. Differentiate to check: $y'=2x+1/x$, which recovers the original differential equation after multiplication by $x$.

Key takeaways

• A differential equation relates an unknown function to its derivatives; solving it means finding a function that satisfies that relationship.
• Order comes from the highest derivative, while degree requires a polynomial equation in derivatives and concerns that highest derivative's power.
• A general solution contains arbitrary constants; a particular solution fixes them using specified information such as an initial point.
• For separable equations, group each variable with its own differential, integrate, and check any values excluded during division.
• A homogeneous equation depends on a ratio of variables; the substitution requires the product rule because the new variable varies.
• For linear equations, normalise first, calculate the integrating factor from the signed coefficient, and integrate the resulting product derivative.
• Treating the other variable as dependent may reveal a linear equation, but the coefficient functions and integration variable must change together.
• Verify a particular solution twice: differentiate to check the equation, then substitute the given values to check its constant.

Test yourself

Why does $y''+2y'+\sin y=0$ have a defined degree?

The equation is polynomial in its derivatives. The sine acts on the dependent variable, so the degree is $1$.

What distinguishes a particular solution from a general solution?

A particular solution fixes the arbitrary constants at specified values; a general solution retains those constants as free parameters.

What must be checked before dividing a separable equation by a function of the dependent variable?

Check where that function is zero and whether any excluded value satisfies the original equation and the supplied condition.

What derivative follows from the homogeneous substitution $y=vx$?

The product rule gives $dy/dx=v+x\,dv/dx$, because the substitution variable $v$ depends on $x$.

What is the integrating factor for $y'-y=\cos x$?

The coefficient of the dependent variable is $-1$, so an integrating factor is $e^{-x}$.

For $dx/dy+P_1(y)x=Q_1(y)$, which variable is used in the integrating-factor integral?

Integrate with respect to $y$, the independent variable in this form; the integrating factor is $e^{\int P_1(y)\,dy}$.

What two checks are needed after finding a particular solution?

Substitute the function and its derivatives into the equation, then confirm that the function satisfies the supplied initial condition.

What differential equation represents a positive principal $P$ growing continuously at $5\%$ per year, with time $t$ measured in years?

The rate equation is $dP/dt=P/20$; the positive amount allows division by the principal during separation.