Answer to Question #156400 in Differential Equations for mohsal

Question

d^2y/dx^2+9y=3+sin (3x)

Answer 1

$Solution:~This ~is~ second~ order~ linear~ non-homogeneous ~differential ~\\ equation~ with ~ constant ~ coefficient ~which ~is ~in ~the ~form~ay''+by'+cy=g(x)$

$\therefore Substitute \frac{d^2y}{dx^2 }=y'' \\ \therefore ~the ~given ~differential ~equation~ is ~written ~as~y''+9y=3+sin(3x)$

$General ~ Solution~ to ~a(x)y''+ b(x)y'+ c(x)y=g(x)~can ~be ~written ~ as~ \\y=y_h +y_p ............................................................................................(1) \\Where~ y_h ~is~ the ~solution~ to~ the~ homogenous ~ODE ~a(x)y''+ b(x)y'+ c(x)y=0 \\and~y_p~particular ~solution, is ~any~ function~ that ~satisfies~the ~\\non-homogenous~equation$

$Now, we ~find, y_h~ and~ y_p .$

$To ~find~ y_h: We~know ~that~, the~ general~ solution~ of~ y''+k^2y=0~is~ \\y=c_1~cos(kx)+c_2~sin(kx) \\\therefore The ~general ~ solution ~ of~ y''+9y=0~\Rightarrow y''+3^2y=0~is~ ~ \\y=c_1~cos(3x)+c_2~sin(3x) \\\therefore y_h=c_1~cos(3x)+c_2~sin(3x).............................................................(2)$

$To ~ find~y_p:~Find ~y_p~that~satisfies~y''+9y=3+sin(3x)\\\Rightarrow y=\frac{1}{3}-\frac{x~cos(3x)}{6} \\ \therefore y_p=\frac{1}{3}-\frac{x~cos(3x)}{6}........................................................................(3)$

$\therefore The~ general ~ solution~ y=y_h+y_p~ is~ \\y=c_1~cos(3x)+c_2~sin(3x)+\frac{1}{3}-\frac{x~cos(3x)}{6}$