Question

Let x = x(y) be the solution of the differential equation
$2ye^{\frac{x}{y^2}}dx+(y^2−4xe^{\frac{x}{y^2}})dy=0$
such that x(1) = 0. Then, x(e) is equal to

• A. e loge(2)
• B. -e loge(2)
• C. e2 loge(2)
• D. -e2 loge(2)

Answer 1

Answer: (D)

-e2 loge(2)

Answer 2

The correct answer is (D) : -e2 loge(2)
Given differential equation
$2ye^{\frac{x}{y^2}}dx+(y^2−4xe^{\frac{x}{y^2}})dy=0 ,x(1)=0$
$⇒e^{\frac{x}{y^2}}[2ydx−4xdy]=−y^2dy$
$⇒e^{\frac{x}{y^2}}[\frac{2ydx−4xdy}{y^4}]=\frac{−1}{y}dy$
$⇒2e^{\frac{x}{y^2}}d(\frac{x}{y^2})=\frac{−1}{y}dy$
$⇒2e^{\frac{x}{y^2}}=$ −ln ⁡y+c…(i)
Now, using x(1) = 0, c = 2
So, for x(e), Put y = e in (i)
$2e^{\frac{x}{e^2}}=−1+2$
$⇒\frac{x}{e^2}$ =ln$⁡(\frac{1}{2})$ ⇒x(e)= −e2ln⁡2