Question

The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is

Options

• x e^y + x² = C

• x e^y + y² = C

• y e^x + x² = C

• y e^y + x² = C

Answer 1

y e^x + x² = C

 

We have,

e^x dy + (ye^x + 2x) dx = 0

\[\text{ Dividing both sides by }e^x dx, \text{ we get }\]

\[\frac{dy}{dx} + \left( y + \frac{2x}{e^x} \right) = 0\]

\[ \Rightarrow \frac{dy}{dx} + y = - \frac{2x}{e^x}\]

\[\text{ Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get }\]

\[P = 1\]

\[Q = - \frac{2x}{e^x}\]

Now,

\[I . F . = e^{\int dx = e^x} \]

Solution is given by,

\[y \times I . F . = \int\left( Q \times I . F . \right) dx + C\]

\[ \Rightarrow y e^x = - \int e^x \times \frac{2x}{e^x}dx + C\]

\[ \Rightarrow y e^x = - 2\int x dx + C\]

\[ \Rightarrow y e^x = - x^2 + C\]

\[ \Rightarrow y e^x + x^2 = C \]