What is the general solution of the differential equation ey′ = x? (A)𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑐 (B) 𝑦 = 𝑥𝑙𝑜𝑔𝑥

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Q. What is the general solution of the differential equation ey' = x? (A)𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑐 (B) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 - 𝑥 + 𝑐 (C) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑥 + 𝑐 (D) 𝑦 = 𝑥 + 𝑐

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Concept understanding — Initial Value Problem

Initial Value Problem

The intuition: a rule of change plus a starting point

A car's speed at time $t$ is $v(t)=\dfrac{ds}{dt}=2t$ . Can you say where the car is at $t=5$ ? Not yet — you don't know where it started ( $0$ m? $10$ m? $100$ m?). The differential equation gives the rule of change, but you also need one starting snapshot to pin down the actual motion. Supply " $s=5$ when $t=0$ " and now everything is determined.

That pairing — a differential equation together with an initial condition — is an Initial Value Problem (IVP).

Note

On its own, a differential equation usually has infinitely many solutions (a whole family of curves, one per value of the arbitrary constant). The initial condition selects exactly one of them.

The precise statement

An IVP has two parts:

A differential equation, e.g. first-order: $\dfrac{dy}{dt}=f(t,y)$ .
An initial condition, the value at a starting point: $y(t_0)=y_0$ .

Written together,

\frac{dy}{dt}=f(t,y),\qquad y(t_0)=y_0,

and the goal is the particular function $y(t)$ satisfying both.

A worked example

Solve $\dfrac{dy}{dt}=3y,\ \ y(0)=2$ .

First solve the equation, ignoring the condition. Separating and integrating, $\dfrac{dy}{y}=3\,dt$ gives $\log|y|=3t+C$ , so the general solution is $y=Ae^{3t}$ . Now apply $y(0)=2$ : $\,2=Ae^{0}=A$ . Hence the unique solution is

y(t)=2e^{3t}.

Watch out

Apply the initial condition last. First get the general solution with its arbitrary constant, then substitute the initial values to fix that constant. Plugging the condition in too early loses solutions and invites algebra errors.

In short: the differential equation says how things change, the initial condition says where they start, and the two together deliver one definite prediction.

Initial value problems are the central focus of the NCERT Class 12 Differential Equations chapter, where finding a 'particular solution' from a 'general solution' plus a given condition is one of the most consistently tested CBSE board question types. Students searching 'initial value problem differential equations class 12' or 'general solution vs particular solution examples' will find this apply-the-condition-last method is exactly the standard board-exam procedure.

The key idea is to rewrite the given differential equation $e y' = x$ as $y' = \frac{x}{e}$ and then integrate directly. The general solution is $y = \frac{x^2}{2e} + C$ , which does not match any of the given options (A)–(D). The options appear to be for a different problem.

The first thing to notice is that the equation is not $e^y y' = x$ or $e^{y'} = x$ — it is simply $e y' = x$ , where $e$ is the constant (Euler's number, approximately 2.718). This is a first-order ordinary differential equation of the simplest kind: the derivative is given explicitly as a function of $x$ .

Why does this matter? Because when $y'$ is expressed directly in terms of $x$ , the solution is just an antiderivative. There is no need for separation of variables, integrating factors, or any special technique. The constant $e$ is just a multiplier.

Let’s work through it.

Rewrite the equation Start with

e y' = x.

Since $e \neq 0$ , divide both sides by $e$ :

y' = \frac{x}{e}.

Interpret $y'$ as $\frac{dy}{dx}$ We have

\frac{dy}{dx} = \frac{x}{e}.

This tells us that the rate of change of $y$ with respect to $x$ is a linear function of $x$ (with slope $1/e$ ).

Integrate both sides with respect to $x$

y = \int \frac{x}{e} \, dx = \frac{1}{e} \int x \, dx.

The integral of $x$ is $\frac{x^2}{2}$ , so

y = \frac{1}{e} \cdot \frac{x^2}{2} + C = \frac{x^2}{2e} + C,

where $C$ is an arbitrary constant of integration.

Check the result Differentiate: $y' = \frac{2x}{2e} = \frac{x}{e}$ . Multiply by $e$ : $e y' = x$ . It works.

Watch out

A common mistake is to misread the equation as $e^y y' = x$ or $e^{y'} = x$ , which would lead to entirely different solutions involving logarithms. The given equation is $e y' = x$ — the constant $e$ times the derivative, not an exponential of $y$ or $y'$ .

Now compare with the options:

(A) $y = x \log x + c$
(B) $y = x \log x - x + c$
(C) $y = x \log x + x + c$
(D) $y = x + c$

None of these match $y = \frac{x^2}{2e} + C$ . The options contain $\log x$ terms, which typically arise from integrating $\frac{1}{x}$ or from solving $y' = \frac{1}{x}$ . That suggests the options belong to a different problem — perhaps $x y' = 1$ or $y' = \frac{1}{x}$ .

Tip

If the intended equation were $e^{y'} = x$ , then taking natural logs gives $y' = \log x$ , and integrating yields $y = x \log x - x + C$ , which is option (B). But that is not what is written. Always read the notation carefully: $e y'$ means $e \cdot y'$ , not $e^{y'}$ .

Since the problem as stated has a clear, correct solution that is absent from the list, the most accurate conclusion is that none of the given options is correct for $e y' = x$ .

✓Final answer

y = \frac{x^2}{2e} + C

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx}=\cos x$
(i) $y=\sin x-2$
(ii) $y=\sin x+1$
(iii) $y=\sin x+2$
(iv) $y=\sin x-1$
›Reveal solutionSolution

Integrate directly and use the initial condition to find $C$ .

$\frac{dy}{dx} = \cos x \implies y = \int \cos x\,dx = \sin x + C$

Apply the initial condition $x=0,\ y=2$ :

$2 = \sin 0 + C = 0 + C \implies C = 2$

So the particular solution is:

$y = \sin x + 2$

✓Final answer
$y=\sin x+2$
CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx}+x=0$
(a) $y+\dfrac{x^2}{2}+1=0$
(b) $y+\dfrac{x^2}{2}=1$
(c) $y+2x^2+1=0$
(d) $y+2x^2=1$
›Reveal solutionSolution

Separate variables, integrate, and use the initial condition $x=0,y=1$ to fix the constant.

$\dfrac{dy}{dx}=-x \;\Rightarrow\; dy=-x\,dx \;\Rightarrow\; \int dy=-\int x\,dx$

$y=-\dfrac{x^2}{2}+C$

At $x=0,y=1$ : $1=0+C\Rightarrow C=1$ .

So $y=-\dfrac{x^2}{2}+1$ , i.e. $y+\dfrac{x^2}{2}=1$ .

✓Final answer
$y+\dfrac{x^2}{2}=1$
CBSE 2024Set ANNUAL1 markMCQ
$\dfrac{d}{dx}f(x) = 4x^3 - \dfrac{3}{x^4}$
(a) $x^4 + \dfrac{1}{x^3} - \dfrac{129}{8}$
(b) $x^3 + \dfrac{1}{x^4} + \dfrac{129}{8}$
(c) $x^4 - \dfrac{1}{x^3} + \dfrac{129}{8}$
(d) $x^3 + \dfrac{1}{x^4} - \dfrac{129}{8}$
›Reveal solutionSolution

Integrate $f'(x)$ term by term, then use $f(2)=0$ to find the constant of integration.

Given $\dfrac{d}{dx}f(x) = 4x^3 - \dfrac{3}{x^4} = 4x^3 - 3x^{-4}$ .

Integrating:

$f(x) = \int (4x^3 - 3x^{-4})\,dx = x^4 - 3\cdot\frac{x^{-3}}{-3} + C = x^4 + x^{-3} + C = x^4 + \frac{1}{x^3} + C$

Using $f(2)=0$ :

$16 + \frac{1}{8} + C = 0 \implies C = -16 - \frac{1}{8} = -\frac{128+1}{8} = -\frac{129}{8}$

So:

$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$

✓Final answer
$x^4+\dfrac{1}{x^3}-\dfrac{129}{8}$
CBSE 2023Set 65/2/11 markMCQ
$\frac{d}{dx}[f(x)] = ax + b$
(a) $a + b$
(b) $\frac{ax^2}{2} + bx$
(c) $\frac{ax^2}{2} + bx + c$
(d) $b$
›Reveal solutionSolution
Integrate the derivative and use the initial condition $f(0) = 0$ to determine the constant; $f(x) = \frac{ax^2}{2} + bx$ .

When you know the derivative of a function, you can recover the original function through integration. The process introduces an arbitrary constant of integration, which you then pin down using any given initial or boundary condition.

Here we're told that $\frac{d}{dx}[f(x)] = ax + b$ , which means the rate of change of $f$ is a linear function of $x$ . To find $f(x)$ itself, we integrate both sides with respect to $x$ .

Finding $f(x)$ by integration

Integrate the derivative.

$f(x) = \int (ax + b) \, dx$
Applying the power rule term by term:
$f(x) = a \cdot \frac{x^2}{2} + bx + C$
where $C$ is the constant of integration that appears whenever we perform an indefinite integral.

Apply the initial condition $f(0) = 0$ .

Substitute $x = 0$ into the expression we just found:

$f(0) = a \cdot \frac{0^2}{2} + b \cdot 0 + C = C$
Since we're given that $f(0) = 0$ , we have:
$C = 0$

Write the final form of $f(x)$ .

With $C = 0$ , the function becomes:

$f(x) = \frac{ax^2}{2} + bx$
Watch out

Option (c) includes an arbitrary constant $c$ , which would be correct for the general antiderivative. But the problem gives us $f(0) = 0$ , which forces that constant to be zero. Don't leave undetermined constants when you have enough information to find them.

✓Final answer
$\frac{ax^2}{2} + bx$

Q.What is the general solution of the differential equation ey′ = x? (A)𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑐 (B) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 − 𝑥 + 𝑐 (C) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑥 + 𝑐 (D) 𝑦 = 𝑥 + 𝑐

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