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 log_e(2)
• B. -e log_e(2)
• C. e² log_e(2)
• D. -e² log_e(2)

Answer 1

The Correct Option is D

The correct answer is (D) : -e² log_e(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