Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

For auxiliary equation,

$D^{2}+D=0$

Solving we get,

$D=-1,0$

$\therefore \mathrm{C.F.}=C_{1} e^{-x}+C_{2} e^{0 x}$

$\therefore C \cdot F=C_{1} e^{-x}+C_{2}$

For $\mathrm{P.I} .$

$y=\frac{e^{4 x}}{D^{2}+D}$

Now, put $D=4$

$\therefore y=\frac{e^{4 x}}{4^{2}+4}=\frac{e^{4 x}}{20}$

$\therefore$ The complete solution is,

$y=C_{1} e^{-x}+C_{2}+\frac{e^{4 x}}{20}$