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

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