Question

The general solution of the differential equation y (x²y + e^x)dx – e^xdy = 0 is –

• A. x³y – 3e^x = cy
• B. x³y + 3e^x = cy
• C. y³x – 3e^y = cx
• D. y³x + 3e^y = cx

Answer 1

Answer: (B)

x³y + 3e^x = cy

Answer 2

We have y (x²y + e^x) dx – e^x dy = 0

Ž e^x $\frac{dy}{dx}$ = x²y² + ye^x

Dividing by y²e^x, we get $\frac{dy}{dx}$ – $\frac{1}{y}$ = x²e^–x

Put $\frac{1}{y}$ = V so that $\frac { - 1 } { y ^ { 2 } }$ $\frac{dy}{dx}$ = $\frac{dV}{dx}$.

We thus have $\frac{dV}{dx}$ + V = –x²e^–x, which is linear

\ I. F. = $e^{\int_{}^{}{1dx}}$= e^x.

Hence the solution is

V . e^x = – $\int_{}^{}{x^{2}e^{–x}}$. e^x dx + $\frac{C}{3}$

or $\frac{1}{y}$e^x = – $\frac{x^{3}}{3}$ + $\frac{C}{3}$ or x³y + 3e^x = Cy