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Find the general solution for each of the following differential equations. \frac{d y}{d x}+\frac{1}{x} \cdot y=x^{2}
CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal
Find the general solution for each of the following differential equations. \frac{d y}{d x}+\frac{1}{x} \cdot y=x^{2}

Solution:

Given Differential Equation :
\frac{\text { dy }}{\mathrm{dx}}+\frac{1}{\mathrm{x}} \cdot \mathrm{y}=\mathrm{x}^{2} \ldots \ldots \ldots \mathrm{eq}(1)
Formula :
i) \int \frac{1}{x} d x=\log x
ii) \int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}
iii) a^{\log _{a} b}=b
iv) General solution :
For the differential equation in the form of
\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}
The general solution is given by,
\text { y.(I.F.) }=\int \text { Q. (I.F.) } \mathrm{dx}+\mathrm{c}
Where integrating factor,
\text { I. } F .=e^{\int P d x}
Answer:
Equation (1) is of the form
\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}
\text { Where, } P=\frac{1}{x} \text { and } Q=x^{2}
Therefore, integrating factor is
\begin{array}{l} \text { I. F. }=\mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}} \\ =\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}} \\ =\mathrm{e}^{\log \mathrm{x}} \ldots \ldots \ldots\left(\because \int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}=\log \mathrm{x}\right) \\ =\mathrm{x} \ldots \ldots \ldots\left(\because \mathrm{a}^{\log _{\mathrm{a}} \mathrm{b}}=\mathrm{b}\right) \end{array}
General solution is
\begin{array}{l} \mathrm{y} \cdot(\mathrm{I} . \mathrm{F} .)=\int \mathrm{Q} \cdot(\mathrm{I} \cdot \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{c} \\ \therefore \mathrm{y} \cdot(\mathrm{x})=\int \mathrm{x}^{2} \cdot(\mathrm{x}) \mathrm{d} \mathrm{x}+\mathrm{c} \\ \therefore \mathrm{xy}=\int \mathrm{x}^{3} \mathrm{~d} \mathrm{x}+\mathrm{c} \\ \therefore \mathrm{xy}=\frac{\mathrm{x}^{4}}{4}+\mathrm{c} \cdots \cdots \cdots\left(\because \int \mathrm{x}^{\mathrm{n}} \mathrm{d} \mathrm{x}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}\right) \\ \therefore y=\frac{x^{3}}{4}+\frac{c}{x} \end{array}

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