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प्रश्न
Solve the following differential equation:-
\[\frac{dy}{dx} + 3y = e^{- 2x}\]
उत्तर
We have,
\[\frac{dy}{dx} + 3y = e^{- 2x} \]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = 3\]
\[Q = e^{- 2x} \]
Now,
\[I . F . = e^{\int P\ dx} \]
\[ = e^{3\int dx} \]
\[ = e^{3x} \]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow y e^{3x} = \int e^{3x} \times e^{- 2x} dx + C\]
\[ \Rightarrow y e^{3x} = e^x + C\]
\[ \Rightarrow y = e^{- 2x} + C e^{- 3x}\]
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