The general solution of $\frac{dy}{dx}=2x\,e^{x^2-y}$ is: (A) $e^{x^2-y}=c$ (B) $e^{-y}+e^{x^2}=c$ (C) $e^y=e^{x^2}+c$ (D) $e^{x^2}+y=c$

The general solution of (dy)/(dx)=2xex^2-y is: (A) ex^2-y=c (B) e-y+ex^2=c (C) ey=ex^2+c (D) ex^2+y=c

NCERT Exemplar · Q83

\frac{dy}{dx}=2x\,e^{x^2-y}

Punjab PsebMCQ· 1mImportance★★★★★

91% · 203/222 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

The temperature difference between the coffee and the room (a function of time)
The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

\frac{dy}{dx} = f(x) \cdot g(y)

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). This moves all $y$ 's to one side and all $x$ 's to the other.

Step 2: Integrate. Put an integral sign on both sides:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

Start your 14-day free trial to unlock the full solution →

This is a first-order ODE that becomes separable after rewriting the right-hand side using exponent rules. The general solution is $e^y = e^{x^2} + c$ , which corresponds to option (C).

The equation is $\frac{dy}{dx} = 2x \, e^{x^2 - y}$ . At first glance, it looks like it might need an integrating factor — but the key is to notice the exponent $x^2 - y$ . Using the law $e^{a-b} = e^a / e^b$ , we can split it:

\frac{dy}{dx} = 2x \, \frac{e^{x^2}}{e^y}.

Now multiply both sides by $e^y$ :

e^y \frac{dy}{dx} = 2x \, e^{x^2}.

The left-hand side is exactly $\frac{d}{dx}(e^y)$ by the chain rule, because $\frac{d}{dx}(e^y) = e^y \frac{dy}{dx}$ . So the equation becomes:

\frac{d}{dx}(e^y) = 2x \, e^{x^2}.

This is now a direct integration problem — no need for an integrating factor at all.

Integrate both sides with respect to $x$ :

\int \frac{d}{dx}(e^y) \, dx = \int 2x \, e^{x^2} \, dx.

The left side gives $e^y$ (plus a constant). For the right side, let $u = x^2$ , so $du = 2x \, dx$ , and the integral becomes $\int e^u \, du = e^u + c = e^{x^2} + c$ .

Write the result: $e^y = e^{x^2} + c.$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ82 — Family y=Ax+A³ of curves will correspond to a differential equation of…→Next questionQ84 — The curve for which the slope of the tangent at any point is equal to …

Q.The general solution of $\frac{dy}{dx}=2x\,e^{x^2-y}$ is:
(A) $e^{x^2-y}=c$
(B) $e^{-y}+e^{x^2}=c$
(C) $e^y=e^{x^2}+c$
(D) $e^{x^2}+y=c$

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Method: Separate variables by splitting an exponential

Steps

Common Mistakes

Q.The general solution of dydx=2x ex2−y\frac{dy}{dx}=2x\,e^{x^2-y}dxdy​=2xex2−y is: (A) ex2−y=ce^{x^2-y}=cex2−y=c (B) e−y+ex2=ce^{-y}+e^{x^2}=ce−y+ex2=c (C) ey=ex2+ce^y=e^{x^2}+cey=ex2+c (D) ex2+y=ce^{x^2}+y=cex2+y=c

🎓Unlock everything free for 14 days

Q.The general solution of $\frac{dy}{dx}=2x\,e^{x^2-y}$ is:
(A) $e^{x^2-y}=c$
(B) $e^{-y}+e^{x^2}=c$
(C) $e^y=e^{x^2}+c$
(D) $e^{x^2}+y=c$

Unlock everything free for 14 days