Differential Equations | ISC Class 12 Maths Notes

Differential equations, order and degree, general and particular solutions, formation by eliminating arbitrary constants, separation of variables, equations reducible to separable form, homogeneous equations, integrating factors, linear equations in either variable, and solution checks.

What is a differential equation?

Definition: A differential equation is an equation involving derivatives of a dependent variable with respect to one or more independent variables. An ordinary differential equation involves derivatives with respect to just one independent variable.

How do the variables and symbols work?

Usually, x is the independent variable and y is the dependent variable, whose value is determined by x. The notation y = f(x) means that the function f assigns a value y to each allowed value of x.

The derivative dy/dx measures the rate at which y changes with x. It is also written y′. On a curve, dy/dx gives the slope of the tangent, the line that has the curve's direction at the point under consideration.

For example, x(dy/dx) + y = 0 involves both a variable and a derivative. In contrast, x + y = 7 relates variables without involving a derivative. The presence of a derivative is the important distinction when recognising a differential equation.

A solution is a function that satisfies the differential equation when its value and the required derivative are substituted. Solving therefore means finding a function, rather than simply finding a numerical value of an unknown.

What does the notation in a calculation mean?

The symbols dx and dy are differential notation used when separating variables or integrating. The symbol ∫ denotes an indefinite integral, which finds an antiderivative: a function whose derivative equals the integrand, the expression being integrated.

Here C denotes an arbitrary constant, a fixed number whose value has not been specified. The symbol e is the base of natural logarithms; log denotes the natural logarithm throughout. The notation |x| means the absolute value of x, and ≠ means “is not equal to”.

The trigonometric symbols sin, cos and tan mean sine, cosine and tangent respectively. Angles in calculus are measured in radians. The domain is the set of allowed values, so division by a variable requires attention to where that variable is nonzero.

How do order and degree differ?

The order of a differential equation is the order of its highest derivative. The first derivative dy/dx has order one. Every solving method developed here concerns equations of first order, so the first derivative is the highest derivative appearing in those equations.

Definition: The degree of a differential equation, when it is polynomial in its derivatives, is the highest positive integer power of its highest-order derivative.

A polynomial in derivatives uses non-negative integer powers of derivatives, with coefficients that may involve the variables. Thus a sine of y does not itself prevent degree from being defined. A sine of dy/dx does prevent the displayed equation from being polynomial in derivatives.

How should the definition be applied?

1. Identify the derivative of highest order in the equation.
2. Check whether the equation is polynomial in its derivatives.
3. If degree is defined, read the highest power of that highest-order derivative.
4. Give order and degree separately; if degree is undefined, say so explicitly.

Worked example 1. Find the order and degree of dy/dx − cos x = 0.

Answer: The highest derivative is dy/dx, so the order is 1. It occurs to the first power and the equation is polynomial in the derivative, so the degree is 1. The trigonometric expression involves x, not the derivative.

In the table, sin² y means (sin y)². The degree condition concerns derivatives, not whether the whole equation is a polynomial in x and y. Order and degree, when defined, are positive integers; an undefined degree should not be reported as zero.

What are general and particular solutions?

A general solution, also called a primitive, contains arbitrary constants. For the first-order equations considered here, the general solution contains one arbitrary constant. Different values of that constant describe different members of a family of solution curves.

A particular solution is obtained by assigning a particular value to the arbitrary constant. It contains no arbitrary constant. An initial condition gives the value of the dependent variable at a specified value of the independent variable and can determine this constant.

How is a proposed solution verified?

Differentiate the proposed function, substitute the function and its derivative into the original equation, and simplify. The two sides must agree throughout the part of the domain being considered. If an initial condition is supplied, check that condition separately.

Worked example 2. Verify that y = x² + 2x + C satisfies y′ − 2x − 2 = 0, where C is an arbitrary constant.

Answer: Differentiating gives y′ = 2x + 2 because the derivative of C is zero. Substitution gives (2x + 2) − 2x − 2 = 0. Thus the proposed family satisfies the equation for every value of C.

The equation above illustrates why verification and solution are different tasks. Verification begins with a proposed function and checks it. Solving begins with the differential equation and uses a method to find a family of functions that satisfies it.

A solution can be explicit, with the dependent variable isolated, or implicit, with the variables related without isolating one of them. An implicit relation can be verified by differentiating both sides and comparing the result with the original differential equation.

The graph of a solution is a solution curve or integral curve. A condition that the curve passes through a point supplies both coordinates: substitute the point's x-coordinate for x and its y-coordinate for y to determine the constant.

How is a differential equation formed by eliminating a constant?

Formation reverses the direction of solving. Start with a family of curves containing an arbitrary constant, differentiate with respect to the independent variable, and remove the constant using the original relation and its derivative. The resulting equation relates the variables and derivatives.

Elimination means removing the arbitrary constant from the relation. It does not mean choosing a convenient numerical value for it. Choosing a value would select one curve, whereas the differential equation must describe the derivative relation shared by the family.

What changes when the constant is additive?

Worked example 3. Form the differential equation of y = x² + 2x + C by eliminating the arbitrary constant C.

Answer: Differentiate with respect to x to obtain dy/dx = 2x + 2. The additive constant disappears on differentiation. Hence the required differential equation is dy/dx − 2x − 2 = 0, of order 1 and degree 1.

What changes when the constant is a coefficient?

Worked example 4. Eliminate the arbitrary constant A from the family y = Ax, for x ≠ 0.

Answer: Differentiating gives dy/dx = A. From the original family, A = y/x. Equating these expressions gives dy/dx = y/x, or x(dy/dx) = y. The constant A has been removed, leaving an equation of order 1 and degree 1.

In the second example, A is the constant slope of an individual straight line. Its value can vary from one member of the family to another, but it remains constant while differentiating any one member. Treating A as a function would change the problem.

Check formation by substituting the original family into the final equation. For y = Ax, the derivative A makes x(dy/dx) = y become xA = Ax. This confirms the elimination without introducing a new condition or selecting a particular line.

How are variables separated and integrated?

A separable equation can be written as dy/dx = g(x)h(y), where g is a function of x alone and h is a function of y alone. Separation places all factors involving y with dy and all factors involving x with dx.

Result: the separated integral relation

Where h(y) ≠ 0, rewrite the equation as dy/h(y) = g(x) dx. Integration gives ∫dy/h(y) = ∫g(x) dx. Equivalently, H(y) = G(x) + C, where H and G are antiderivatives of 1/h(y) and g(x), respectively.

One arbitrary constant is sufficient after integrating both sides. If separate constants are initially written on each side, their difference can be combined into one arbitrary constant. The symbols representing a constant can also be renamed after multiplication by a fixed nonzero number.

Worked example 5. Solve dy/dx = (x + 1)/(2 − y), where y ≠ 2.

Answer: Separate to get (2 − y) dy = (x + 1) dx. Integration gives 2y − y²/2 = x²/2 + x + C₁, where C₁ is an arbitrary constant. Rearranging gives x² + y² + 2x − 4y + C = 0, with C = 2C₁.

This is an implicit general solution. There is no need to solve the quadratic relation for y to obtain a valid integrated answer. Differentiating gives 2x + 2y(dy/dx) + 2 − 4(dy/dx) = 0, which recovers the given equation where y ≠ 2.

Worked example 6. Solve dy/dx = (1 + y²)/(1 + x²) for real x and y.

Answer: Separate to obtain dy/(1 + y²) = dx/(1 + x²). Integration gives tan⁻¹ y = tan⁻¹ x + C. Here tan⁻¹ means the inverse tangent function, not the reciprocal of tangent. Both denominators are positive for real variables.

The variables have been separated completely only when each integral contains its own variable. A remaining factor involving y in the x-integral, or involving x in the y-integral, means the separation step is unfinished.

How are initial conditions and curve conditions used?

Find the general solution first, then use the supplied condition to determine its constant. Keep the original equation's restrictions while doing this. A point on a curve and an initial value both supply numerical values that can be substituted into the integrated relation.

Worked example 7. Solve dy/dx = −4xy², given y = 1 when x = 0.

Answer: For y ≠ 0, separate to obtain dy/y² = −4x dx. Integration gives −1/y = −2x² + C. Substituting x = 0 and y = 1 gives C = −1. Hence the particular solution is y = 1/(2x² + 1).

The final function gives y = 1 at x = 0. Its derivative is −4x/(2x² + 1)², which equals −4xy² after substituting the function for y. These are two distinct checks: the differential equation and the initial condition must both be satisfied.

Note: Dividing by y² assumes y ≠ 0. The constant function y = 0 also satisfies the original equation in Worked example 7, but it does not satisfy the supplied initial condition y = 1 at x = 0.

How does a tangent condition become an equation?

Worked example 8. Find the curve through (−2, 3) whose tangent slope at (x, y) is 2x/y², with y ≠ 0.

Answer: Write dy/dx = 2x/y². Separation gives y² dy = 2x dx, so y³/3 = x² + C. Substituting (−2, 3) gives 9 = 4 + C, hence C = 5. The required curve is y³/3 = x² + 5.

The slope condition supplies the differential equation; the given point selects a particular solution. Neither part replaces the other. Without the point, the answer remains a family containing C. Without the slope relation, the point alone does not determine this curve.

Notice that the slope formula itself excludes y = 0. This restriction comes from the denominator of the original expression, so it must be considered even when a rearranged or integrated equation no longer displays that denominator.

How can an equation be reduced to separable form?

Some equations become separable after a substitution, meaning a new variable is introduced to represent an expression involving the original variables. The substitution must also be differentiated. Replacing an expression without transforming its derivative does not produce an equivalent equation.

How does a repeated combination guide substitution?

Worked example 9. Find the particular solution of (x − y)(dx + dy) = dx − dy, given y = −1 when x = 0. Use t = x − y, where t is a new dependent variable.

Answer: Dividing by dx gives (x − y)(1 + y′) = 1 − y′. Since t = x − y, dt/dx = 1 − y′, so y′ = 1 − dt/dx.

Substitution gives t(2 − dt/dx) = dt/dx, hence 2t = (t + 1)dt/dx. For t ≠ 0, separate as (1 + 1/t) dt = 2 dx.

Integrating gives t + log|t| = 2x + C. The initial values give t = 1 and C = 1. Replacing t by x − y yields x − y + log|x − y| = 2x + 1.

The combination x − y appears in both the equation and the supplied hint. Its derivative produces the combination 1 − y′, making the substitution effective. The transformed problem now has one variable with dt and the other with dx.

Always substitute back before presenting an answer in the original variables. A relation involving t alone is an intermediate result unless the problem specifically asks for the transformed equation. Check the original initial values after the substitution has been reversed.

The division by t excludes t = 0 during the calculation. That case corresponds to y = x, which satisfies the original differential relation but not the given point. The required solution has t = 1 initially, so the logarithmic calculation applies near that point.

How are homogeneous differential equations recognised and solved?

A function F(x, y) is homogeneous of degree n if F(λx, λy) = λⁿF(x, y), where n is its degree of homogeneity and λ is a nonzero scaling constant. This test compares what happens when both variables are scaled together.

For example, y² + 2xy is homogeneous of degree two, while 2x − 3y is homogeneous of degree one. The function cos(y/x) has degree zero where defined, because scaling both variables leaves their ratio unchanged. This degree concerns a function, not powers of derivatives.

Result: substitution for a homogeneous equation

A differential equation dy/dx = F(x, y) is homogeneous when F has degree zero. Write it as dy/dx = g(y/x), where g is a function of the ratio. For x ≠ 0, put v = y/x, so y = vx.

Here v is a new dependent variable, so the product rule gives dy/dx = v + x(dv/dx). Consequently x(dv/dx) = g(v) − v. Where g(v) − v ≠ 0, integrate dv/[g(v) − v] = dx/x and then replace v by y/x.

Worked example 10. Find the family of curves with dy/dx = (x² + y²)/(2xy), where x and y are nonzero.

Answer: Divide numerator and denominator by x² to obtain dy/dx = [1 + (y/x)²]/[2(y/x)]. Put y = vx. Then v + x(dv/dx) = (1 + v²)/(2v).

Rearrangement gives x(dv/dx) = (1 − v²)/(2v). Separating and integrating gives −log|1 − v²| = log|x| + C₁. Combining logarithms and renaming the constant gives x(1 − v²) = C.

Replace v by y/x and multiply by x to obtain x² − y² = Cx. The excluded cases v = 1 and v = −1 give y = x and y = −x; both are included by C = 0.

The constant-ratio cases deserve checking because division by 1 − v² removed them from the integration step. In this example they are valid solutions on intervals avoiding x = 0 and already belong to the final family.

For an equation written as dx/dy = h(x/y), where h is a function of that ratio, use x = vy instead. Now v depends on y and dx/dy = v + y(dv/dy). The choice of substitution follows the derivative's independent variable.

How does an integrating factor solve a linear equation?

Definition: A first-order linear differential equation has the form dy/dx + Py = Q, where P and Q are constants or functions of x alone. The dependent variable and its derivative occur linearly.

First make the coefficient of dy/dx equal to one. Only then identify P, including its sign, and Q. If a nonzero variable coefficient multiplies the derivative, divide the entire equation by it, keeping the restriction under which that division is valid.

Result: the integrating-factor formula

An integrating factor, abbreviated I.F., is a multiplying function that makes the left side a product derivative. Write this function as μ(x), where μ is the Greek letter mu. Take μ = e^(∫P dx); the notation e^(...) places the enclosed expression in the exponent.

1. Multiply dy/dx + Py = Q by μ to obtain μ(dy/dx) + μPy = μQ.
2. Differentiate μ = e^(∫P dx), giving dμ/dx = Pμ.
3. Use the product rule: d(μy)/dx = μ(dy/dx) + y(dμ/dx) = μ(dy/dx) + μPy.
4. Integrate to obtain μy = ∫μQ dx + C, and divide by μ if an explicit expression for y is wanted.

Worked example 11. Solve x(dy/dx) + 2y = x², where x ≠ 0.

Answer: Divide by x to get dy/dx + (2/x)y = x. Thus P = 2/x and Q = x. An integrating factor is x², because d(x²)/dx = (2/x)x².

Multiply through to obtain x²(dy/dx) + 2xy = x³, or d(x²y)/dx = x³. Integration gives x²y = x⁴/4 + C. Hence y = x²/4 + C/x², for x ≠ 0.

The normalisation step is essential: using 2 as P before dividing by x would produce the wrong integrating factor. Likewise, multiplying only the y-term by the factor would destroy the equality. Apply the factor to the derivative term and the right side too.

An arbitrary multiplicative nonzero constant in an integrating factor does not change the solution family. It is therefore sufficient to choose one convenient factor and keep the arbitrary constant in the final integrated relation.

How are linear equations in x solved, and how is a method chosen?

Some equations are simpler when x is the dependent variable and y is independent. The linear form is dx/dy + Px = Q, where P and Q are now constants or functions of y alone. Integrations must then be performed with respect to y.

Result: the linear formula with variables reversed

For dx/dy + Px = Q, choose μ = e^(∫P dy). The solution is μx = ∫μQ dy + C. This is the same product-derivative argument as before, with the roles of x and y exchanged throughout.

Worked example 12. Solve y dx − (x + 2y²) dy = 0, treating x as a function of y on an interval where y ≠ 0.

Answer: Rewrite as dx/dy − x/y = 2y. Thus P = −1/y and Q = 2y. Choose μ = 1/y; its derivative with respect to y is −1/y² = Pμ.

Multiplication gives (1/y)(dx/dy) − x/y² = 2, or d(x/y)/dy = 2. Integrating gives x/y = 2y + C. Therefore x = 2y² + Cy, on the interval considered.

What features identify the method?

After choosing a method, preserve the original equation for verification. A rearranged equation may have been obtained by division, which can remove possible solutions. The final check should therefore include the derivative relation, the supplied point if any, and restrictions introduced along the way.

Keep the distinction between a general and a particular answer visible. A general solution retains C; a particular solution uses the supplied data to fix it. No value for C should be invented when the question gives no condition.

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 whose derivatives are taken with respect to just one independent variable.
• Order — The order of the highest derivative appearing in the differential equation under consideration.
• Degree — The highest power of the highest-order derivative, defined when the equation is polynomial in derivatives.
• General solution — A solution containing arbitrary constants, with one arbitrary constant for the first-order equations considered here.
• Particular solution — A solution obtained by fixing the arbitrary constants, so that no arbitrary constant remains.
• Solution curve — The graph of a function satisfying the differential equation, also called an integral curve.
• Initial condition — A specified dependent-variable value at a given independent-variable value, used to determine a solution constant.
• Separable equation — An equation allowing factors involving each variable to be grouped with that variable's differential.
• Homogeneous function — A function whose value scales by a fixed power when both arguments are scaled together.
• Linear differential equation — A first-order equation expressible as dy/dx + Py = Q, with P and Q depending on x alone.
• Integrating factor — A function multiplying a linear equation so its left side becomes the derivative of a product.

Common errors and misconceptions

• Misconception: Order and degree are interchangeable. Correct: Order identifies the highest derivative; degree concerns its power after checking polynomial dependence on derivatives.
• Misconception: Any trigonometric expression makes degree undefined. Correct: Distinguish a trigonometric function of a variable from a trigonometric function of a derivative.
• Misconception: Forming an equation means assigning a value to its constant. Correct: Differentiate and eliminate the constant while retaining the relation shared by the family.
• Misconception: From y = vx, dy/dx = x(dv/dx). Correct: The product rule gives dy/dx = v + x(dv/dx), since both factors vary with x.
• Misconception: A variable factor can be cancelled without further thought. Correct: Check its zero cases in the original equation; division may have removed valid solutions.
• Misconception: Read P before making the derivative's coefficient one. Correct: Divide the entire equation first, then identify P and Q with their signs.
• Misconception: Every integrating factor uses integration with respect to x. Correct: For an equation linear in x with derivative dx/dy, integrate with respect to y.
• Misconception: A particular solution may retain an undetermined constant. Correct: Use all supplied initial data to fix the constant and check the resulting function.

Exam-style questions with model answers

Q1. Find the order and degree, if defined, of dy/dx + sin(dy/dx) = 0, where y depends on x. [2 marks]

1. The order is one because the highest derivative present is the first derivative dy/dx.
2. The degree is undefined because the sine of the derivative makes this equation non-polynomial in its derivative.

Q2. Eliminate the arbitrary constant A from y = Ax, where x ≠ 0, to form its differential equation. [3 marks]

1. Differentiate the family with respect to x. Since A is constant within each member of the family, this gives dy/dx = A.
2. Use the original relation y = Ax to express the same constant as A = y/x, which is valid because x is nonzero.
3. Equate the two expressions for A: dy/dx = y/x. Thus the required differential equation is x(dy/dx) = y.

Q3. Find the general solution of dy/dx = (x + 1)/(2 − y), where y ≠ 2. [3 marks]

1. Separate the variables by multiplying through by 2 − y and using differential notation: (2 − y) dy = (x + 1) dx.
2. Integrate both sides to obtain 2y − y²/2 = x²/2 + x + C₁, where C₁ is an arbitrary integration constant.
3. Multiply by two and rearrange. Renaming 2C₁ as C gives the implicit general solution x² + y² + 2x − 4y + C = 0.

Q4. Solve dy/dx = −4xy², given that y = 1 when x = 0. [4 marks]

1. Near the initial point y is nonzero, so separation gives dy/y² = −4x dx.
2. Integrate both sides to obtain −1/y = −2x² + C, where C is an arbitrary constant.
3. Substitute the given values x = 0 and y = 1. This gives −1 = C.
4. Therefore y = 1/(2x² + 1). Its derivative is −4x/(2x² + 1)², equal to −4xy², and it gives y = 1 at x = 0.

Q5. Solve dy/dx = (x² + y²)/(2xy), where x and y are nonzero, using the homogeneous substitution. Include any constant-ratio solutions removed during division. [5 marks]

1. Dividing numerator and denominator by x² gives [1 + (y/x)²]/[2(y/x)], a function of y/x alone. Thus the equation is homogeneous.
2. Set v = y/x, so y = vx and dy/dx = v + x(dv/dx). Substitution gives x(dv/dx) = (1 − v²)/(2v).
3. For v² ≠ 1, separate and integrate: 2v dv/(1 − v²) = dx/x, giving −log|1 − v²| = log|x| + C₁.
4. Combine the logarithms, rename the constant and substitute v = y/x. The resulting general solution is x² − y² = Cx.
5. The divided-out cases v = 1 and v = −1 yield y = x and y = −x. Both satisfy the original equation where x ≠ 0 and are included by C = 0.

Q6. Find the general solution of x(dy/dx) + 2y = x², where x ≠ 0, using an integrating factor. [5 marks]

1. Divide every term by x to obtain dy/dx + (2/x)y = x. This is linear, with coefficient P = 2/x and right side Q = x.
2. Choose the integrating factor μ = x². It satisfies dμ/dx = Pμ because 2x = (2/x)x² on the specified domain.
3. Multiply the entire linear equation by x². The result is x²(dy/dx) + 2xy = x³, whose left side is d(x²y)/dx.
4. Integrate with respect to x to get x²y = x⁴/4 + C, where C is an arbitrary constant.
5. Divide by x² to obtain y = x²/4 + C/x². The condition x ≠ 0 remains necessary for this expression and the preceding divisions.

Q7. Solve y dx − (x + 2y²) dy = 0 as a linear equation for x in terms of y, on an interval where y ≠ 0. [4 marks]

1. Rearrange to dx/dy − x/y = 2y, so P = −1/y and Q = 2y in the linear form.
2. Choose the integrating factor 1/y, whose derivative −1/y² equals the product of P and 1/y.
3. Multiplying gives d(x/y)/dy = 2. Integrating with respect to y yields x/y = 2y + C.
4. Multiply by y to obtain x = 2y² + Cy, with C arbitrary, on the interval where the division by y is valid.

Q8. Find the particular solution of (x − y)(dx + dy) = dx − dy, given y = −1 at x = 0. Use t = x − y. [5 marks]

1. Divide by dx to write (x − y)(1 + y′) = 1 − y′. The supplied substitution gives dt/dx = 1 − y′.
2. Replace y′ by 1 − dt/dx to obtain t(2 − dt/dx) = dt/dx, or 2t = (t + 1)dt/dx.
3. The initial values give t = 1. For the nonzero t branch, separate as (1 + 1/t) dt = 2 dx and integrate.
4. Integration gives t + log|t| = 2x + C. At x = 0, t = 1, so C = 1.
5. Substitute back to obtain x − y + log|x − y| = 2x + 1. The omitted t = 0 solution is y = x and fails the supplied initial condition.

Key takeaways

• A differential equation relates a dependent variable and its derivatives; its solution is a function satisfying that relation.
• Order concerns the highest derivative, while degree requires polynomial dependence on derivatives before its power can be identified.
• A first-order general solution retains an arbitrary constant; a particular solution fixes that constant using supplied data.
• Form a differential equation by differentiating a family and eliminating its arbitrary constant without choosing a particular curve.
• Separation groups each variable with its differential; check zero cases before dividing by variable factors.
• For homogeneous equations involving y/x, substitute y = vx and use the complete product-rule derivative.
• Normalise a linear equation before identifying its coefficients, then apply an integrating factor to every term.
• Verify the final derivative relation, any initial condition, and domain restrictions inherited from the original equation.

Test yourself

What distinguishes an ordinary differential equation?

Its derivatives are taken with respect to just one independent variable.

Why does dy/dx + sin(dy/dx) = 0 have undefined degree?

The derivative occurs inside sine, so the equation is not polynomial in its derivative.

What is the difference between forming and solving a differential equation?

Formation eliminates constants from a family by differentiation; solving finds functions satisfying a given derivative relation.

If v = y/x and y depends on x, what is dy/dx?

Since y = vx, the product rule gives dy/dx = v + x(dv/dx).

What must be checked before dividing a separated equation by h(y)?

Check where h(y) is zero, because constant solutions may be lost through that division.

For dy/dx + Py = Q, with P and Q functions of x alone, what integrating factor is used?

Use e^(∫P dx), after making the coefficient of dy/dx equal to one.

For dx/dy + Px = Q, with P and Q functions of y alone, which variable is used in integration?

Integrate with respect to y; the integrating factor is e^(∫P dy).

Why should a solution satisfying an initial condition still be differentiated and checked?

Passing through the supplied point does not establish that the function satisfies the differential equation throughout its domain.