Subject : Applied Mathematics 2

Topic : Linear Differential Equations With Constant Coefficients and Variable Coefficients Of Higher Order

Difficulty: High

Solution:

put $\frac{d}{d x}=D$

$\left(D^{2}+1\right) y=\frac{1}{1+\sin x}$

For complementary solution,

$f(D)=0$

$\therefore\left(D^{2}+1\right)=0$

Roots are $: \mathrm{D}=\boldsymbol{i},-\boldsymbol{i}$

Roots of given diff. eqn are complex.

The complementary solution of given diff. eqn is given by,

$\therefore y_{c}=c_{1} \cos x+c_{2} \sin x$

For particular solution,

By method of variation of parameters,

$\quad y_{p}=y_{1} p_{1}+y_{2} p_{2}$

$\begin{aligned} \text { where } \boldsymbol{p}_{\mathbf{1}} &=\int \frac{-\boldsymbol{y}_{2} \boldsymbol{X}}{\boldsymbol{w}} d x \\ \boldsymbol{p}_{2} &=\int \frac{\boldsymbol{y}_{1} \boldsymbol{X}}{\boldsymbol{w}} d \boldsymbol{x} \\ \boldsymbol{w} &=\left| \begin{array}{cc}{\boldsymbol{y}_{1}} & {\boldsymbol{y}_{2}} \\ {\boldsymbol{y}_{1}^{\prime}} & {\boldsymbol{y}_{2}^{\prime}}\end{array}\right| \end{aligned}$

$\begin{aligned} w &=\left| \begin{array}{cc}{\cos x} & {\sin x} \\ {-\sin x} & {\cos x}\end{array}\right|=\cos ^{2} x+\sin ^{2} x=1 \\ p_{1} &=\int \frac{-y_{2} X}{w} d x \\ &=\int-\frac{\sin x}{1} \cdot \frac{1}{1+\sin x} d x \\ &=-\int \frac{\sin x}{1+\sin x} \frac{(1-\sin x)}{1+\sin x} \frac{(1-\sin x)}{1+\sin x} \frac{(1-\sin x)}{1+\sin x} d x \\ &=-\int\left(\sec x . \tan x-\tan ^{2} x\right) d x \\ &=-[\sec x-\tan x+x] \end{aligned}$

$\boldsymbol{p}_{2}=\int \frac{\boldsymbol{y}_{1} \cdot \boldsymbol{X}}{\boldsymbol{w}} \boldsymbol{d} \boldsymbol{x}=\int \frac{\cos x}{1} \frac{1}{1+\sin x} \boldsymbol{d} \boldsymbol{x}=\log (\mathbf{1}+\sin \boldsymbol{x})$

$y_{p}=-[\sec x-\tan x+x] \cos x+\log (1+\sin x) \sin x$

The general solution of given diff. eqn is given by,

$ \mathbf{ y_{g}=y_{c}+y_{p} \\ =c_{1} \cos x+c_{2} \sin x-[\sec x-\tan x+x] \cos x+ log(1 + sinx) sinx }$