Differential EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Differential Equation Solution: x(e^2) = 2e^2/3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let x=x(y)x=x(y) be the solution of the differential equation 2y2dxdy−2xy+x2=02y^2\dfrac{dx}{dy}-2xy+x^2=0, y>1y>1, x(e)=ex(e)=e. Then x(e2)x(e^2) is equal to
A32e2\dfrac32 e^2
B23e2\dfrac23 e^2correct
Ce2e^2
D2e22e^2
Solution
Step 1: Write the equation in differential form:
2y(y dx−x dy)+x2 dy=0.2y(y\,dx-x\,dy)+x^2\,dy=0.
 Step 2: Divide by x2x^2 and use d ⁣(yx)=x dy−y dxx2d\!\left(\dfrac{y}{x}\right)=\dfrac{x\,dy-y\,dx}{x^2}, i.e. y dx−x dyx2=−d ⁣(yx)\dfrac{y\,dx-x\,dy}{x^2}=-d\!\left(\dfrac{y}{x}\right):
−2y d ⁣(yx)+dy=0.-2y\,d\!\left(\frac{y}{x}\right)+dy=0.
 Step 3: Divide by yy:
−2 d ⁣(yx)+1y dy=0.-2\,d\!\left(\frac{y}{x}\right)+\frac{1}{y}\,dy=0.
 Step 4: Integrate:
−2yx+log⁡ey=C.-\frac{2y}{x}+\log_e y=C.
 Step 5: Apply x(e)=ex(e)=e (so y=e, x=ey=e,\ x=e): −2+1=C⇒C=−1-2+1=C\Rightarrow C=-1. Thus
2yx−log⁡ey=1.\frac{2y}{x}-\log_e y=1.
 Step 6: Put y=e2y=e^2:
2e2x−2=1 ⇒ 2e2x=3 ⇒ x=2e23.\frac{2e^2}{x}-2=1\ \Rightarrow\ \frac{2e^2}{x}=3\ \Rightarrow\ x=\frac{2e^2}{3}.
 Correct answer: (2)
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