written 5.0 years ago by | • modified 5.0 years ago |
Subject : Applied Mathematics 2
Topic : Linear Differential Equations With Constant Coefficients and Variable Coefficients Of Higher Order
Difficulty: High
written 5.0 years ago by | • modified 5.0 years ago |
Subject : Applied Mathematics 2
Topic : Linear Differential Equations With Constant Coefficients and Variable Coefficients Of Higher Order
Difficulty: High
written 5.0 years ago by |
Solution:
$\left(D^{2}+2\right) y=e^{x} \cos x+x^{2} e^{3 x}$
For complementary solution,
$f(D)=0$
$\therefore\left(D^{2}+2\right)=0$
Roots are : $D=\sqrt{2} i,-\sqrt{2} i$
Roots of given diff. eqn are complex.
The complementary solution of given diff. eqn is given by,
$\quad \therefore y_{c}=c_{1} \cos \sqrt{2} x+c_{2} \sin \sqrt{2} x$
$\begin{aligned} \text { For particular integral, } \\ y_{p} =\frac{1}{f(D)} X=\frac{1}{D^{2}+1} e^{x} \cos x+\frac{1}{D^{2}+1} x^{2} e^{3 x} \\ =e^{x} \frac{1}{(D+1)^{2}+1} \cos x+\frac{1}{D^{2}+1} x^{2} e^{3 x} \\ =e^{x} \frac{1}{D^{2}+2 D+3} \cos x+e^{3 x} \frac{1}{(D+3)^{2}+2} x^{2} \end{aligned}$
$\begin{aligned} &=e^{x} \frac{1}{2} \frac{D-1}{D^{2}-1} \cdot \cos x+e^{3 x} \frac{1}{D^{2}+6 D+11} x^{2} \\ &=e^{x} \frac{1}{4}(\sin x+\cos x)+\frac{e^{3 x}}{11}\left[1+\frac{6 D+D^{2}}{11}\right]^{-1} x^{2} \\ &=e^{x} \frac{1}{4}(\sin x+\cos x)+\frac{e^{3 x}}{11}\left[1-\frac{6 D+D^{2}}{11}+\frac{36 D^{2}}{121}+. .\right] x^{2} \\ \therefore y_{p} &=e^{x} \frac{1}{4}(\sin x+\cos x)+\frac{e^{3 x}}{11}\left[x^{2}-\frac{12 x}{11}+\frac{50}{121}\right] \end{aligned}$
The general solution of given diff. eqn is,
$y_{g}=y_{c}+y_{p}=c_{1} \cos \sqrt{2} x+c_{2} \sin \sqrt{2} x+e^{x} \frac{1}{4}(\sin x+\cos x)+\frac{e^{3 x}}{11}\left[x^{2}-\frac{12 x}{11}+\frac{50}{121}\right]$