Answer to Question #290999 in Differential Equations for CHINTU

Question

Solve (d ^ 2 * y)/(d * x ^ 2) - 2tan x * (dy)/(dx) + 5y =

Answer 1

We are given

$y''-2\tan x(y')+5y=e^x \sec x$

$\cos xy''-2\sin x(y')+5y\cos x=e^x$

We will be looking for a solution in the form

$u(x)=y(x)\cos x$

Then

$u'=y'\cos x-y\sin x$

$u''=y''\cos x-2y'\sin x-y\cos x$

Hence

$\cos xy''-2\sin x(y')=u''+u$

Substitute

$u''+u+5u=e^x$

Corresponding homogeneous differentional equation

$u''+6u=0$

Characteristic (auxiliary) equation

$r^2+6=0$

$r=\pm i\sqrt{6}$

The general solution of the homogeneous differentional equation is

$u_h=c_1\cos (\sqrt{6}x)+c_2 \sin (\sqrt{6}x)$

Find the partial solution in the form

$u_p=Ae^x$

$u_p'=Ae^x$

$u_p''=Ae^x$

Substitute

$Ae^x+6Ae^x=e^x$

$A=\dfrac{1}{7}$

$u_p=\dfrac{1}{7}e^x$

The general solution of the non homogeneous differentional equation is

$u=c_1\cos (\sqrt{6}x)+c_2 \sin (\sqrt{6}x)+\dfrac{1}{7}e^x$

Therefore the solution of the given differential equation is

$y=\dfrac{1}{\cos x}(c_1\cos (\sqrt{6}x)+c_2 \sin (\sqrt{6}x)+\dfrac{1}{7}e^x)$