Question

9y''-4y=sinx

Answer 1

Сharacteristic equation:

$9{k^2} - 4 = 0$

${k^2} = \frac{4}{9}$

$k = \pm \frac{2}{3}$

Then the general solution of the homogeneous equation is

${y_0} = {C_1}{e^{\frac{2}{3}x}} + {C_2}{e^{ - \frac{2}{3}x}}$

We will seek a particular solution in the form

$\widetilde y = A\sin x + B\cos x \Rightarrow {\widetilde y^\prime } = A\cos x - B\sin x \Rightarrow {\widetilde y^{\prime \prime }} = - A\sin x - B\cos x$

Substitute the obtained values ​​into the original equation:

$- 9A\sin x - 9B\cos x - 4A\sin x - 4B\cos x = \sin x$

$\left\{ \begin{array}{l} - 13A = 1\\ - 13B = 0 \end{array} \right. \Rightarrow A = - \frac{1}{{13}},\,B = 0$

So,

$\widetilde y = - \frac{1}{{13}}\sin x$

$y = {y_0} + y = {C_1}{e^{\frac{2}{3}x}} + {C_2}{e^{ - \frac{2}{3}x}} - \frac{1}{{13}}\sin x$

Answer: $y = {C_1}{e^{\frac{2}{3}x}} + {C_2}{e^{ - \frac{2}{3}x}} - \frac{1}{{13}}\sin x$