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Find the General Solution of the Differential Equation X D Y D X + 2 Y = X 2 - Mathematics

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प्रश्न

Find the general solution of the differential equation \[x\frac{dy}{dx} + 2y = x^2\]

योग
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उत्तर

We have,
\[ x\frac{dy}{dx} + 2y = x^2 \]
\[ \Rightarrow \frac{dy}{dx} + \frac{2}{x}y = x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = \frac{2}{x}\text{ and }Q = x . \]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\frac{2}{x} dx} \]
\[ = e^{2\log x} \]
\[ = x^2 \]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = x^2 ,\text{ we get }\]
\[ x^2 \left( \frac{dy}{dx} + \frac{2}{x}y \right) = x^2 x \]
\[ \Rightarrow x^2 \frac{dy}{dx} + 2xy = x^3 \]
Integrating both sides with respect to x, we get
\[ x^2 y = \int x^3 dx + C\]
\[ \Rightarrow x^2 y = \frac{x^4}{4} + C\]
\[ \Rightarrow y = \frac{x^2}{4} + C x^{- 2} \]
\[\text{ Hence, }y = \frac{x^2}{4} + C x^{- 2} \text{ is the required solution . }\]

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अध्याय 22: Differential Equations - Exercise 22.10 [पृष्ठ १०७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.10 | Q 38 | पृष्ठ १०७

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