For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of

Miscellaneous Exercise · Q2

Q. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i)

xy = a e^x + b e^{-x} + x^2

x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0

(ii)

y = e^x (a \cos x + b \sin x)

\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0

(iii)

y = x \sin 3x

\frac{d^2 y}{dx^2} + 9y - 6 \cos 3x = 0

(iv)

x^2 = 2y^2 \log y

(x^2 + y^2) \frac{dy}{dx} - xy = 0

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Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

A function $y = \phi(x)$ is called a solution of a differential equation if, when you substitute it and its derivatives into the equation, the two sides become equal for every $x$ in the domain. Verification is the act of carrying out that substitution and checking that it holds as an identity.

The useful point: you do not have to solve the equation to verify a candidate. You are only checking a function that is already handed to you — which is exactly how many exam questions are phrased: "Show that $\dots$ is a solution of $\dots$ ."

The steps

From the given $y = \phi(x)$ , compute exactly the derivatives that appear in the equation.
Substitute $y$ and those derivatives into the left-hand side.
Simplify and check that it equals the right-hand side for all $x$ (an identity, not just at one point).

Example 1

Verify that $y = e^{-3x}$ is a solution of $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$ .

Here $y' = -3e^{-3x}$ and $y'' = 9e^{-3x}$ . Substituting:

9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} = (9 - 3 - 6)e^{-3x} = 0.

The left side is $0$ for every $x$ , so $y = e^{-3x}$ is a solution.

Example 2 (a solution with constants)

Verify that $y = a\cos x + b\sin x$ satisfies $\dfrac{d^2y}{dx^2} + y = 0$ for any constants $a, b$ .

Since $y'' = -a\cos x - b\sin x = -y$ , we get $y'' + y = 0$ . It holds for all $a, b$ , so this two-constant family is a solution.

Note

Verification links your answer back to the definition of a solution: a function is a solution not because of how you found it, but because it makes the differential equation true. If the substitution does not reduce to an identity, the function is simply not a solution.

Verifying that a given function solves a differential equation is explicitly listed as an exercise type in the NCERT Class 12 Mathematics textbook's Differential Equations chapter, and "verify the solution of differential equation examples" is a common CBSE and JEE Main search. This is often the easiest full-mark question in the chapter once the substitution steps are practiced a few times.

Concept: Verification of Solution — substitute the given function into the differential equation and check that it satisfies the identity.

(i) $xy = a e^x + b e^{-x} + x^2$

Differentiate twice:

$y + x y' = a e^x - b e^{-x} + 2x$

$2y' + x y'' = a e^x + b e^{-x} + 2$

Now $x y'' + 2 y' - x y + x^2 - 2 = (a e^x + b e^{-x} + 2) - x y + x^2 - 2 = (a e^x + b e^{-x} + x^2) - x y = 0$ since $x y = a e^x + b e^{-x} + x^2$ . Verified.

(ii) $y = e^x (a \cos x + b \sin x)$

$y' = e^x (a \cos x + b \sin x) + e^x (-a \sin x + b \cos x) = y + e^x (-a \sin x + b \cos x)$

$y'' = y' + e^x (-a \sin x + b \cos x) + e^x (-a \cos x - b \sin x) = y' + (y' - y) - y = 2y' - 2y$

Thus $y'' - 2y' + 2y = 0$ . Verified.

(iii) $y = x \sin 3x$

$y' = \sin 3x + 3x \cos 3x$

$y'' = 3 \cos 3x + 3 \cos 3x - 9x \sin 3x = 6 \cos 3x - 9y$

Then $y'' + 9y - 6 \cos 3x = (6 \cos 3x - 9y) + 9y - 6 \cos 3x = 0$ . Verified.

(iv) $x^2 = 2y^2 \log y$

Differentiate implicitly: $2x = 4y \log y \cdot y' + 2y^2 \cdot \frac{1}{y} y' = 4y y' \log y + 2y y'$

So $2x = 2y y' (2 \log y + 1)$ → $y' = \frac{x}{y (2 \log y + 1)}$

Now $(x^2 + y^2) y' - x y = \left(2y^2 \log y + y^2\right) \cdot \frac{x}{y (2 \log y + 1)} - x y = \frac{y^2 (2 \log y + 1) \cdot x}{y (2 \log y + 1)} - x y = x y - x y = 0$ . Verified.

✓Final answer

All four given functions are verified as solutions of their respective differential equations.

For each part, we verify the given function satisfies the differential equation by computing the required derivatives, substituting them into the equation, and simplifying to an identity (0 = 0). The key is careful differentiation — product rule, chain rule, and implicit differentiation where needed.

(i) $xy = a e^x + b e^{-x} + x^2$ : $x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$

Concept: The given relation is implicit in $y$ . We can either solve for $y$ explicitly or differentiate the equation as it stands. Since $y$ appears multiplied by $x$ , solving explicitly is straightforward: $y = \frac{a e^x + b e^{-x} + x^2}{x}$ . Then we compute $y'$ and $y''$ and substitute.

Write $y$ explicitly:

y = \frac{a e^x}{x} + \frac{b e^{-x}}{x} + x

First derivative (using quotient rule on the first two terms, or rewrite as $a e^x x^{-1}$ and differentiate):

y' = a\left(\frac{e^x}{x} - \frac{e^x}{x^2}\right) + b\left(-\frac{e^{-x}}{x} - \frac{e^{-x}}{x^2}\right) + 1

Factor common terms:

y' = \frac{a e^x (x-1)}{x^2} - \frac{b e^{-x} (x+1)}{x^2} + 1

Second derivative: Differentiate $y'$ term by term. For the first term, use quotient rule on $\frac{a e^x (x-1)}{x^2}$ :
- Let $u = a e^x (x-1)$ , $v = x^2$ . Then $u' = a e^x (x-1) + a e^x = a e^x x$ , and $v' = 2x$ .
- So derivative = $\frac{(a e^x x)(x^2) - (a e^x (x-1))(2x)}{x^4} = \frac{a e^x x^3 - 2a e^x x(x-1)}{x^4} = \frac{a e^x (x^2 - 2x + 2)}{x^3}$
For the second term, $-\frac{b e^{-x} (x+1)}{x^2}$ :
- $u = -b e^{-x}(x+1)$ , $v = x^2$ . $u' = -b[-e^{-x}(x+1) + e^{-x}] = -b e^{-x}(-x) = b e^{-x} x$ (careful: derivative of $e^{-x}(x+1)$ is $-e^{-x}(x+1) + e^{-x} = -x e^{-x}$ , so $u' = -b(-x e^{-x}) = b x e^{-x}$ ).
- Derivative = $\frac{(b x e^{-x})(x^2) - (-b e^{-x}(x+1))(2x)}{x^4} = \frac{b e^{-x} x^3 + 2b e^{-x} x(x+1)}{x^4} = \frac{b e^{-x} (x^2 + 2x + 2)}{x^3}$
The derivative of $1$ is $0$ . So:

y'' = \frac{a e^x (x^2 - 2x + 2)}{x^3} + \frac{b e^{-x} (x^2 + 2x + 2)}{x^3}

Substitute into the differential equation: $x y'' + 2 y' - x y + x^2 - 2 = 0$

Compute $x y''$ :

x y'' = \frac{a e^x (x^2 - 2x + 2)}{x^2} + \frac{b e^{-x} (x^2 + 2x + 2)}{x^2}

Compute $2 y'$ :

2 y' = \frac{2a e^x (x-1)}{x^2} - \frac{2b e^{-x} (x+1)}{x^2} + 2

Compute $-x y$ :

-x y = -x\left(\frac{a e^x}{x} + \frac{b e^{-x}}{x} + x\right) = -a e^x - b e^{-x} - x^2

Now sum everything:

x y'' + 2 y' - x y + x^2 - 2 = \left[\frac{a e^x (x^2 - 2x + 2)}{x^2} + \frac{b e^{-x} (x^2 + 2x + 2)}{x^2}\right] + \left[\frac{2a e^x (x-1)}{x^2} - \frac{2b e^{-x} (x+1)}{x^2} + 2\right] + \left[-a e^x - b e^{-x} - x^2\right] + x^2 - 2

Combine the $a e^x$ terms: factor $\frac{a e^x}{x^2}$ :

\frac{a e^x}{x^2}[(x^2 - 2x + 2) + 2(x-1)] - a e^x = \frac{a e^x}{x^2}[x^2 - 2x + 2 + 2x - 2] - a e^x = \frac{a e^x}{x^2}[x^2] - a e^x = a e^x - a e^x = 0

Combine the $b e^{-x}$ terms: factor $\frac{b e^{-x}}{x^2}$ :

\frac{b e^{-x}}{x^2}[(x^2 + 2x + 2) - 2(x+1)] - b e^{-x} = \frac{b e^{-x}}{x^2}[x^2 + 2x + 2 - 2x - 2] - b e^{-x} = \frac{b e^{-x}}{x^2}[x^2] - b e^{-x} = b e^{-x} - b e^{-x} = 0

The constant terms: $+2 - x^2 + x^2 - 2 = 0$ .

Everything cancels. Hence the given function satisfies the differential equation.

Watch out

A common mistake is forgetting the $+x$ term when writing $y$ explicitly from $xy = ...$ — that $x$ comes from $x^2/x = x$ , not from the exponential terms. Also, when differentiating $b e^{-x}/x$ , the sign of the derivative of $e^{-x}$ is $-e^{-x}$ , so handle with care.

(ii) $y = e^x (a \cos x + b \sin x)$ : $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$

Concept: This is a linear combination of $e^x \cos x$ and $e^x \sin x$ , which are known to satisfy the second-order linear ODE $y'' - 2y' + 2y = 0$ . We verify by direct differentiation.

First derivative: Use product rule. Let $u = e^x$ , $v = a \cos x + b \sin x$ .

y' = e^x (a \cos x + b \sin x) + e^x (-a \sin x + b \cos x) = e^x[(a+b)\cos x + (b-a)\sin x]

Second derivative: Differentiate $y'$ again. Write $y' = e^x[(a+b)\cos x + (b-a)\sin x]$ . Apply product rule:

y'' = e^x[(a+b)\cos x + (b-a)\sin x] + e^x[-(a+b)\sin x + (b-a)\cos x]

Simplify:

y'' = e^x[(a+b + b-a)\cos x + (b-a - a - b)\sin x] = e^x[2b \cos x - 2a \sin x]

Substitute into $y'' - 2y' + 2y$ :

y'' - 2y' + 2y = e^x[2b \cos x - 2a \sin x] - 2e^x[(a+b)\cos x + (b-a)\sin x] + 2e^x[a \cos x + b \sin x]

Factor $e^x$ and collect $\cos x$ and $\sin x$ terms:

For $\cos x$ : $2b - 2(a+b) + 2a = 2b - 2a - 2b + 2a = 0$

For $\sin x$ : $-2a - 2(b-a) + 2b = -2a - 2b + 2a + 2b = 0$

Hence the expression is identically zero.

Tip

Notice that $y = e^x (a \cos x + b \sin x)$ is the general solution of $y'' - 2y' + 2y = 0$ . The characteristic equation is $r^2 - 2r + 2 = 0$ , with roots $r = 1 \pm i$ , giving exactly this form. So verification is essentially checking that the function matches the known solution form.

(iii) $y = x \sin 3x$ : $\frac{d^2 y}{dx^2} + 9y - 6 \cos 3x = 0$

Concept: Here $y$ is a product of $x$ and $\sin 3x$ . We compute two derivatives and substitute. The presence of $-6\cos 3x$ in the equation suggests that after substitution, the $\sin$ terms will cancel and leave a $\cos$ term that matches.

First derivative: Using product rule:

y' = \sin 3x + x \cdot 3 \cos 3x = \sin 3x + 3x \cos 3x

Second derivative: Differentiate $y'$ :

y'' = 3 \cos 3x + 3 \cos 3x + 3x \cdot (-3 \sin 3x) = 6 \cos 3x - 9x \sin 3x

Substitute into $y'' + 9y - 6 \cos 3x$ :

y'' + 9y - 6 \cos 3x = (6 \cos 3x - 9x \sin 3x) + 9(x \sin 3x) - 6 \cos 3x

The $6 \cos 3x$ and $-6 \cos 3x$ cancel. The $-9x \sin 3x + 9x \sin 3x$ also cancel. Result is $0$ .

Watch out

A common error: forgetting the factor of 3 when differentiating $\sin 3x$ (chain rule). Also, when differentiating $3x \cos 3x$ , the derivative of $\cos 3x$ is $-3 \sin 3x$ , giving $-9x \sin 3x$ , not $-3x \sin 3x$ .

(iv) $x^2 = 2y^2 \log y$ : $(x^2 + y^2) \frac{dy}{dx} - xy = 0$

Concept: This relation is implicit. We differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ , then solve for $\frac{dy}{dx}$ and substitute into the given equation.

Differentiate the given relation implicitly:

\frac{d}{dx}(x^2) = \frac{d}{dx}(2y^2 \log y)

Left side: $2x$ .

Right side: $2 \cdot \frac{d}{dx}(y^2 \log y)$ . Use product rule: derivative of $y^2$ is $2y \frac{dy}{dx}$ , derivative of $\log y$ is $\frac{1}{y} \frac{dy}{dx}$ .

So:

\frac{d}{dx}(y^2 \log y) = 2y \frac{dy}{dx} \cdot \log y + y^2 \cdot \frac{1}{y} \frac{dy}{dx} = 2y \log y \frac{dy}{dx} + y \frac{dy}{dx} = y(2 \log y + 1) \frac{dy}{dx}

Hence:

2x = 2 \cdot y(2 \log y + 1) \frac{dy}{dx} \quad \Rightarrow \quad 2x = 2y(2 \log y + 1) \frac{dy}{dx}

Solve for $\frac{dy}{dx}$ :

\frac{dy}{dx} = \frac{x}{y(2 \log y + 1)}

Substitute into $(x^2 + y^2) \frac{dy}{dx} - xy$ :

(x^2 + y^2) \cdot \frac{x}{y(2 \log y + 1)} - xy

Factor $x$ :

= x \left[ \frac{x^2 + y^2}{y(2 \log y + 1)} - y \right]

Combine inside the bracket over a common denominator:

= x \left[ \frac{x^2 + y^2 - y^2(2 \log y + 1)}{y(2 \log y + 1)} \right] = x \left[ \frac{x^2 + y^2 - 2y^2 \log y - y^2}{y(2 \log y + 1)} \right] = x \left[ \frac{x^2 - 2y^2 \log y}{y(2 \log y + 1)} \right]

Use the original relation: $x^2 = 2y^2 \log y$ . Substitute $x^2$ :

x^2 - 2y^2 \log y = 2y^2 \log y - 2y^2 \log y = 0

Hence the numerator is zero, so the entire expression is zero.

Tip

The key insight: the differential equation is designed so that after substituting $\frac{dy}{dx}$ from the implicit relation, the numerator simplifies to $x^2 - 2y^2 \log y$ , which is exactly zero by the given relation. So the verification reduces to recognizing that the given equation is used to eliminate the numerator.

✓Final answer

All four given functions are verified to be solutions of their respective differential equations.

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Q.For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

The steps

Example 1

Example 2 (a solution with constants)

(i) $xy = a e^x + b e^{-x} + x^2$ : $x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$

(ii) $y = e^x (a \cos x + b \sin x)$ : $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$

(iii) $y = x \sin 3x$ : $\frac{d^2 y}{dx^2} + 9y - 6 \cos 3x = 0$

(iv) $x^2 = 2y^2 \log y$ : $(x^2 + y^2) \frac{dy}{dx} - xy = 0$

Method: Verifying a given function solves a differential equation

Steps

Common Mistakes

Q.For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) xy=aex+be−x+x2xy = a e^x + b e^{-x} + x^2xy=aex+be−x+x2 : xd2ydx2+2dydx−xy+x2−2=0x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0xdx2d2y​+2dxdy​−xy+x2−2=0

(ii) y=ex(acos⁡x+bsin⁡x)y = e^x (a \cos x + b \sin x)y=ex(acosx+bsinx) : d2ydx2−2dydx+2y=0\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0dx2d2y​−2dxdy​+2y=0

(iii) y=xsin⁡3xy = x \sin 3xy=xsin3x : d2ydx2+9y−6cos⁡3x=0\frac{d^2 y}{dx^2} + 9y - 6 \cos 3x = 0dx2d2y​+9y−6cos3x=0

(iv) x2=2y2log⁡yx^2 = 2y^2 \log yx2=2y2logy : (x2+y2)dydx−xy=0(x^2 + y^2) \frac{dy}{dx} - xy = 0(x2+y2)dxdy​−xy=0

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(i) $xy = a e^x + b e^{-x} + x^2$ : $x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$

(ii) $y = e^x (a \cos x + b \sin x)$ : $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$

(iii) $y = x \sin 3x$ : $\frac{d^2 y}{dx^2} + 9y - 6 \cos 3x = 0$

(iv) $x^2 = 2y^2 \log y$ : $(x^2 + y^2) \frac{dy}{dx} - xy = 0$

Unlock everything free for 14 days