Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

$\begin{array}{ll}{\frac{d^{2} y}{d x^{2}}+3 \frac{d y}{d x}+2 y=e^{e^{x}}} \\ {\text { Put } D=\frac{d}{d x}} & {\therefore D^{2} y+3 D y+2 y=0}\end{array}$

For complementary solution,

$f(D)=0$

$\therefore D^{2}+3 D+2=0$

$D=-1,-2$

$\therefore \quad y_{c}=c_{1} e^{-x}+c_{2} e^{-2 x}$

Particular integral is given by,

$\boldsymbol{y}_{p}=\boldsymbol{y}_{1} \boldsymbol{p}_{1}+\boldsymbol{y}_{2} \boldsymbol{p}_{2}$

$\begin{aligned} \text { Where } \boldsymbol{p}_{1} &=\int \frac{-y_{2} X}{w} d x \\ p_{2} &=\int \frac{y_{1} X}{w} d x \\ w &=\left| \begin{array}{ll}{y_{1}} & {y_{2}} \\ {y_{1}^{\prime}} & {y_{2}^{\prime}}\end{array}\right| \end{aligned}$

$\therefore w=\left| \begin{array}{cc}{e^{-x}} & {e^{-2 x}} \\ {-e^{-x}} & {-2 e^{-2 x}}\end{array}\right|=-e^{-3 x}$

$\boldsymbol{p}_{1}=\int \frac{e^{-2 x} \cdot e^{e^{x}}}{e^{-3 x}} \boldsymbol{d} x=\int e^{e^{x}} \cdot e^{x} d x=\int e^{t} d t=e^{e^{x}} \ldots \ldots\left\{ \text{ put } e^{x}=t =\gt e^{x} d x=d t\right\}$

$\boldsymbol{p}_{2}=\int \frac{e^{-x}}{-e^{-3 x}} \cdot e^{e^{x}} d x=\int e^{e^{x}} \cdot e^{2 x} d x=\int t . e^{t} d t=e^{x} e^{e^{x}}-e^{e^{x}}$

$\quad \therefore y_{p}=e^{x} e^{e^{x}}-\left(e^{x} e^{e^{x}}-e^{e^{x}}\right) \cdot e^{-2 x} \cdot e^{-2 x} \cdot e^{e^{x}}$

The general solution of given differential eqn is given by,

$y_{g}=y_{c}+y_{p}=c_{1} e^{-x}+c_{2} e^{-2 x}+e^{-2 x} \cdot e^{e^{x}}$