Answer to Question #293049 in Differential Equations for jacob

Question

 5. If the function y = y(x) is such that x dy dx + 2y = x cos x, use the integrating-factor method to show that y = sin x + 2 cos x /x − 2 sin x/x^2 + c/x^2 , where c is an arbitrary constant

Answer 1

Solution:

$\begin{aligned} &\text { Given } x \frac{d y}{d x}+2 y=x \cos x \\ &\Rightarrow\left(x \frac{d y}{d x}+2 y\right) \times \frac{1}{x}=x \cos x \times \frac{1}{x} \\ &\Rightarrow \frac{d y}{d x}+\frac{2 y}{x}=\cos x \\ &\Rightarrow \frac{d y}{d x}+\left(\frac{2}{x}\right) y=\cos x \end{aligned}$

This is a first order linear differential equation of the form

$\frac{d y}{d x}+P y=Q$  Here, $P=\frac{2}{x}\ and\ Q=\cos x$

The integrating factor (I.F) of this differential equation is,

$\begin{aligned} &\text { I. F }=\mathrm{e}^{\int \mathrm{Pdx}} \\ &\Rightarrow \text { I.F }=\mathrm{e}^{\int \frac{2}{\mathrm{x}} \mathrm{dx}} \\ &\Rightarrow \text { I.F }=\mathrm{e}^{2 \int \frac{1}{\mathrm{x}} \mathrm{dx}} \end{aligned} \\ \begin{aligned} &\text { We have } \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log \mathrm{x}+\mathrm{c} \\ &\Rightarrow \text { I.F }=\mathrm{e}^{2 \log \mathrm{x}} \end{aligned}$

$\begin{aligned} &\Rightarrow \mathrm{I} . \mathrm{F}=\mathrm{e}^{\log \mathrm{x}^{2}} \\ { } \\ {\left[\because \mathrm{m} \log \mathrm{a}=\log \mathrm{a}^{\mathrm{m}}\right]} \\ {\therefore \mathrm{I} . \mathrm{F}=\mathrm{x}^{2}\left[\because \mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\right]} \end{aligned}$

Hence, the solution of the differential equation is,

$\begin{aligned} y(I . F)=& \int(Q \times I . F) d x+c \\ \Rightarrow y\left(x^{2}\right)=& \int\left(\cos x \times x^{2}\right) d x+c \\ \Rightarrow y x^{2}=& \int x^{2} \cos x d x+c \\ \Rightarrow y x^{2}=& \int\left(x^{2}\right) \times(\cos x) d x+c \\ \text { Recall } \int f(x) g(x)=f(x)\left[\int g(x) d x\right] \\ &-\int\left[f^{\prime}(x)\left(\int g(x) d x\right)\right] d x+c \\ \Rightarrow y x^{2}=& x^{2}\left[\int \cos x d x\right] \\ &-\int\left[\frac{d}{d x}\left(x^{2}\right)\left(\int \cos x d x\right)\right] d x+c \end{aligned}$

$\begin{aligned} \Rightarrow y x^{2} &=x^{2} \sin x-2 \int x \sin x d x+c \\ \Rightarrow y x^{2} &=x^{2} \sin x-2\left\{x\left[\int \sin x d x\right]\right.\\ &\left.-\int\left[\frac{d}{d x}(x)\left(\int \sin x d x\right)\right] d x\right\}+c \\ \Rightarrow y x^{2} &=x^{2} \sin x-2\{x[-\cos x]\\ \Rightarrow y x^{2} &=x^{2} \sin x-2\{-x \cos x\\ \Rightarrow y x^{2} &=x^{2} \sin x \\ &-2\{-x \cos x+\sin x\}+c \\ \Rightarrow y x^{2} &=x^{2} \sin x+2 x \cos x \\ & \quad-2 \sin x+c \\ \end{aligned}$

 $\mathrm{y}=\sin \mathrm{x}+\frac{2}{\mathrm{x}} \cos \mathrm{x}-\frac{2}{\mathrm{x}^{2}} \sin \mathrm{x}+\frac{\mathrm{c}}{\mathrm{x}^{2}}$

Thus, the solution of the given differential equation is

$y=\sin x+\frac{2}{x} \cos x-\frac{2}{x^{2}} \sin x+\frac{c}{x^{2}}$