Solve the equation $y^{\prime\prime}+4y=\frac{2}{\sin(2x)}$
I can find the complementary solution $y_c=c_1(\cos(2x))+c_2(\sin(2x))$
But I'm having problems finding the particular solution.
I want to choose something $A\frac{1}{\sin(x)}$ I believe.
Solve the equation $y^{\prime\prime}+4y=\frac{2}{\sin(2x)}$
ordinary differential equations
Best Answer
$$y^{\prime\prime}+4y=\frac{2}{\sin(2x)}\tag 1$$ A particular solution of the associated homogeneous equation is : $y=c \sin(2x)$.
Change of function (The constant $c$ is replaced by an unknown function $u(x)$ : $$y=u(x)\sin(2x)$$ $y'=u'\sin(2x)+2u\cos(2x)$
$y''= u''\sin(2x)+4u'\cos(2x)-4u\sin(2x)$
Equation $(1)$ is transformed into $$u''\sin(2x)+4u'\cos(2x)=\frac{2}{\sin(2x)}$$ Let $u'(x)=v(x)$ $$v'(x)\sin(2x)+4v(x)\cos(2x)=\frac{2}{\sin(2x)}$$ This is a linear first order ODE. You certainly can continue.