Find a particular solution satisfying the given condition for each of the following differential equations. $\mathrm{x} \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}+\mathrm{y}=\mathrm{x}^{3}$, given that $y=1$ when $x=2$

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Find a particular solution satisfying the given condition for each of the following differential equations. $\mathrm{x} \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}+\mathrm{y}=\mathrm{x}^{3}$ , given that $y=1$ when $x=2$

CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal

Find a particular solution satisfying the given condition for each of the following differential equations. $\mathrm{x} \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}+\mathrm{y}=\mathrm{x}^{3}$ , given that $y=1$ when $x=2$

Solution:

$\mathrm{x} \frac{d y}{d x}+y=x^{3}\dots(1)$
General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ is given by,
$y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$
Where, integrating factor,
$\text { I. } F .=e \int^{p d x}$
Dividing equation $(1)$ by $\mathrm{x}$ we get
$\frac{d y}{d x}+\frac{1}{x} \cdot y=x^{2} \ldots \ldots \ldots(2)$
Equation (2) is of the form
$\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
Where, $\mathrm{P}=1 / \mathrm{x}$ and $\mathrm{Q}=\mathrm{x}^{2}$
Therefore, integrating factor is
$\text { I. } F .=e \int^{p d x}=e^{\int \frac{1}{x} d x}=\mathrm{x}$
General solution is
$\begin{array}{l} y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c \\ y \cdot x=\int x^{2} x \mathrm{dx}+\mathrm{c} \\ \mathrm{xy}=\int x^{3} d x+c \\ \mathrm{xy}=\frac{x^{4}}{4}+c \end{array}$
Dividing above equation by $x$ ,
$\mathrm{y}=\frac{x^{3}}{4}+\frac{c}{x} \text { (general equation) }$
For particular solution put $\mathrm{y}=1$ and $\mathrm{x}=2$ in above equation $\mathrm{c}=-2$
Therefore, particular solution is
$\mathrm{y}=\frac{x^{3}}{4}-\frac{2}{x}$