$(D^2 – 7D – 6) y = (1 + x^2) e^2x$

Subject : Applied Mathematics 2

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

Difficulty: High

Solution:

$\left(D^{2}-7 D-6\right) y=\left(1+x^{2}\right) e^{2 x}$

For complementary solution,

$f(D)=0$

$\therefore\left(D^{2}-7 D-6\right)=0$

Roots are: $D=\frac{7}{2}+\frac{\sqrt{73}}{2}, \frac{7}{2}-\frac{\sqrt{73}}{2}$

Roots of the given diff. eqn are irrational roots.

$\boldsymbol{y}_{c}=e^{\frac{7 x}{2}}\left(c_{1} \cosh \frac{\sqrt{73}}{2}+c_{2} \sinh \frac{\sqrt{73}}{2}\right)$

For particular integral,

$\begin{aligned} y_{p} &=\frac{1}{f(D)} X \\ &=\frac{1}{\left(D^{2}-7 D-6\right)}\left[e^{2 x}+e^{2 x} x^{2}\right] \\ &=\frac{1}{\left(D^{2}-7 D-6\right)} e^{2 x}+\frac{1}{\left(D^{2}-7 D-6\right)} e^{2 x} x^{2} \\ &=-\frac{e^{2 x}}{16}+e^{2 x} \frac{1}{(D+2)^{2}-7(D+2)-6} x^{2} \\ &=-\frac{e^{2 x}}{16}+e^{2 x} \frac{1}{D^{2}-3 D-16} x^{2} \\ &=-\frac{e^{2 x}}{16}+e^{2 x}\left[\frac{1}{-16}\left(\frac{1}{1+\frac{3 D-D^{2}}{16}}\right)\right] x^{2} \\ &=-\frac{e^{2 x}}{16}+e^{2 x}\left[\frac{1}{-16}\left(\frac{1}{1+\frac{3 D-D^{2}}{16}}\right)\right] x^{2} \\ &=-\frac{e^{2 x}}{16}\left[1+\left(1+\frac{3 D-D^{2}}{16}\right)^{-1} x^{2}\right] \\ &=-\frac{e^{2 x}}{16}\left[1+\left[1 \frac{3 D-D^{2}}{16}+\left(\frac{3 D-D^{2}}{16}\right)^{2}\right] x^{2}\right\} \\ &=-\frac{e^{2 x}}{16}\left\{1+\left[x^{2}-\frac{3}{8} x+\frac{2}{16}+\frac{9}{16 \times 8}\right]\right\} \\ &=-\frac{e^{2 x}}{16}\left\{1+\left[x^{2}-\frac{3}{8} x+\frac{25}{128}\right]\right\} \\ y_{p} &=-\frac{e^{2 x}}{16}-\frac{e^{2 x}}{16}\left[x^{2}-\frac{3}{8} x+\frac{25}{128}\right] \\ \end{aligned}$

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

$y_{g}=y_{c}+y_{p}=e^{\frac{7 x}{2}}\left(c_{1} \cosh \frac{\sqrt{73}}{2}+c_{2} \sinh \frac{\sqrt{73}}{2}\right)-\frac{e^{2 x}}{16}\left[x^{2}-\frac{3}{8} x+\frac{25}{128}\right]$