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प्रश्न
\[\frac{dy}{dx} + 2y = e^{3x}\]
योग
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उत्तर
We have,
\[\frac{dy}{dx} + 2y = e^{3x} . . . . . (1)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 2\]
\[Q = e^{3x} \]
\[ \therefore I . F . = e^{\int P dx} \]
\[ = e^{\int2 dx} \]
\[ = e^{2x} \]
\[\text{ Multiplying both sides of (1) by }e^{2x} ,\text{ we get }\]
\[ e^{2x} \left( \frac{dy}{dx} + 2y \right) = e^{2x} e^{3x} \]
\[ \Rightarrow e^{2x} \frac{dy}{dx} + 2 e^{2x} y = e^{5x} \]
Integrating both sides with respect to x, we get
\[y e^{2x} = \int e^{5x} dx + C\]
\[ \Rightarrow y e^{2x} = \frac{e^{5x}}{5} + C\]
\[ \Rightarrow y = \frac{1}{5} e^{3x} + C e^{- 2x} \]
\[\text{ Hence, }y = \frac{1}{5} e^{3x} + C e^{- 2x}\text{ is the required solution .} \]
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