Answer to Question #126547 in Differential Equations for Tina

Question

Solve the following differential equations cosx d²y/dx²+sinx dy/dx - 2ycos³x = 2cos^5 x

Answer 1

$\cos xy''+\sin xy'-2y\cos^3x=2\cos^5x$

Using a substitution

$y=-\cos^2x+1+z(x)\\ y'=\sin2x+z'(x)\\ y''=2\cos2x+z''(x)$

Then

$\cos x(2\cos2x+z'')+\sin x(\sin2x+z')-\\ -2(-\cos^2x+1+z)\cos^3x=2\cos^5x\\ \cos xz''+\sin xz'-2\cos^3xz=0$

Using a substitution

$\sin x=t\\ x=\sin^{-1}t\\ z'=\frac{z'_t}{x'_t}=z'\sqrt{1-t^2}\\ z''=\frac{z''_t}{x'_t}=(z''(1-t^2)+z'\frac{-2t}{\sqrt{1-t^2}})\sqrt{1-t^2}\\$

Put it into the equation

$(1-t^2)^{\frac{3}{2}}(z''-2z)=0\\ z''-2z=0\\ \lambda^2-2=0\\ \lambda_1=\sqrt2\\ \lambda_2=-\sqrt2\\ z=C_1e^{\sqrt2 t}+C_2e^{-\sqrt2 t}\\ z=C_1e^{\sqrt2 \sin x}+C_2e^{-\sqrt2 \sin x}\\ y=-\cos^2 x+1+C_1e^{\sqrt2 \sin x}+C_2e^{-\sqrt2 \sin x}\\$