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MCQMediumJEE 2026Homogeneous Differential Equations

JEE Mathematics 2026 Question with Solution

Let the solution curve of the differential equation x dy−y dx=x2+y2 dx,x>0,x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx,\quad x>0, with y(1)=0y(1)=0, be y=y(x)y=y(x). Then y(3)y(3) is equal to

  • A

    44

  • B

    22

  • C

    11

  • D

    66

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The differential equation is

x dy−y dx=x2+y2 dxx\,dy - y\,dx = \sqrt{x^2+y^2}\,dx

with initial condition y(1)=0y(1)=0.

Find: The value of y(3)y(3).

Rewrite the equation in standard form by dividing by dxdx:

xdydx−y=x2+y2x\frac{dy}{dx} - y = \sqrt{x^2+y^2}

Use the substitution y=vxy = vx. Then

y=vx⇒dydx=v+xdvdxy = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}

Substitute into the differential equation:

x(v+xdvdx)−vx=x2+v2x2x\left(v + x\frac{dv}{dx}\right) - vx = \sqrt{x^2 + v^2x^2}

So,

x2dvdx=x1+v2x^2\frac{dv}{dx} = x\sqrt{1+v^2}

Since x>0x>0, divide by xx to get

xdvdx=1+v2x\frac{dv}{dx} = \sqrt{1+v^2}

Hence the variables separate as

dv1+v2=dxx\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}

Integrate both sides:

∫dv1+v2=∫dxx\int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x}

Therefore,

sinh⁡−1(v)=ln⁡x+C\sinh^{-1}(v) = \ln x + C

Apply the initial condition. Since v=yxv=\frac{y}{x}, at x=1x=1 and y=0y=0 we get v=0v=0.

sinh⁡−1(0)=ln⁡1+C\sinh^{-1}(0) = \ln 1 + C

So,

C=0C=0

Thus,

sinh⁡−1(v)=ln⁡x\sinh^{-1}(v)=\ln x

Now express vv in terms of xx:

v=sinh⁡(ln⁡x)=x−1x2v = \sinh(\ln x)=\frac{x-\frac{1}{x}}{2}

Since y=vxy=vx,

y=vx=x2−12y = vx = \frac{x^2-1}{2}

Evaluate at x=3x=3:

y(3)=9−12=4y(3)=\frac{9-1}{2}=4

Therefore, the correct option is A.

Using the homogeneous substitution explicitly

The expression x2+y2\sqrt{x^2+y^2} suggests writing yy as a multiple of xx because the right-hand side becomes a function of yx\frac{y}{x} after factoring out xx.

With y=vxy=vx,

x2+y2=x2+v2x2=x2(1+v2)=x1+v2\sqrt{x^2+y^2} = \sqrt{x^2+v^2x^2} = \sqrt{x^2(1+v^2)} = x\sqrt{1+v^2}

because x>0x>0 is given. This is the key simplification.

Then the equation reduces to a separable form, and after integration the initial condition forces the constant to be zero. Substituting back gives

y=x2−12y=\frac{x^2-1}{2}

so at x=3x=3,

y(3)=4y(3)=4

Hence the correct option is A.

Common mistakes

  • Taking x2+v2x2=∣x∣1+v2\sqrt{x^2+v^2x^2} = |x|\sqrt{1+v^2} and then forgetting to use the condition x>0x>0. Here ∣x∣=x|x|=x, which is necessary to obtain the correct separable equation.

  • Using the substitution y=vxy=vx but writing dydx=xdvdx\frac{dy}{dx}=x\frac{dv}{dx} only. The product rule gives dydx=v+xdvdx\frac{dy}{dx}=v+x\frac{dv}{dx}, and missing the vv term changes the equation completely.

  • Applying the initial condition directly to yy after integrating in terms of vv. Once the equation is integrated in vv, the condition must be converted using v=yxv=\frac{y}{x} first.

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