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प्रश्न
The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
विकल्प
x ey + x2 = C
x ey + y2 = C
y ex + x2 = C
y ey + x2 = C
MCQ
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उत्तर
y ex + x2 = C
We have,
ex dy + (yex + 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,
\[y \times I . F . = \int\left( Q \times I . F . \right) dx + C\]
\[ \Rightarrow y e^x = - \int e^x \times \frac{2x}{e^x}dx + C\]
\[ \Rightarrow y e^x = - 2\int x dx + C\]
\[ \Rightarrow y e^x = - x^2 + C\]
\[ \Rightarrow y e^x + x^2 = C \]
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