Answer to Question #127869 in Differential Equations for Raghisa

Question

d^2y/dx^2-2tanxdy/dx+3y=2secx given that y=sinx is a solution of associated homogeneous equation

Answer 1

y''-2tan(x)y'+3y=2/sec(x) (1)

y=sin(x),y'=cos(x),y''=-sin(x)

y"-2tan(x)y'+3y=-sin(x)-2tan(x)cos(x)+3sin(x)=0

Really, y=sinx is a solution of associated homogeneous equation

y"-2tan(x)y'+3y=0.

Let's find the second solution y₂(x) of the homogeneous equation in the form

y(x)=p(x)sin(x)

y'=p'sin(x)+pcos(x)

y''=p''sin(x)+2p'cos(x)-psin(x)

p''sin(x)+2p'cos(x)-psin(x)-2tan(x)(p'sin(x)+pcos(x))+3psin(x)=0

p''sin(x)+2p'cos(x)-psin(x)-2p'sin²(x)/cos(x)-2psin(x)+3psin(x)=0

p''sin(x)+2p'cos(x)-2p'sin²(x)/cos(x)=0

z=p'

z'sin(x)+2zcos(x)-2zsin²2(x)/cos(x)=0

z'sin(x)=-2zcos(2x)/cos(x)

dz/z=-2cos(2x)/(cos(x)sin(x))dx

dz/z=-4cos(2x)/sin(2x)dx

$\int$ dz/z=-4$\int$cos(2x)/sin(2x)dx

ln|z|=-2ln|sin(2x)|+C₁

z=C₂/sin²(2x)

p'=C₂/sin²(2x)

p(x)=-C₂cot(2x)/2

Let C₂=-4

p(x)=2cot(2x)

y(x)=2cot(2x)sin(x)

y(x)=2cos(2x)sin(x)/sin(2x)

y(x)=2cos(2x)sin(x)/(2sin(x)cos(x))

y(x)=cos(2x)/cos(x)

So,

y₂(x)=cos(2x)/cos(x)

The general solution of the homogeneous equation has the form

y(x)=C₁sin(x)+C₂cos(2x)/cos(x)

The equation (1) can be transform into standart form using the substitution

ln(y)=ln(z)-1/2$\int$(-2tan(x))dx

ln(y)=ln(z)-ln(cos(x))

y=z/cos(x)

z=cos(x)y

z'=-sin(x)y+cos(x)y'

z''=-cos(x)y-2sin(x)y'+cos(x)y''

cos(x)(y"-2tan(x)y'+3y)=cos(x)y''-2sin(x)y'+3cos(x)y=z''+4z

Multiplying the equation (1) by cos(x) we obtain the equation

z''+4z=2

We will seek a particular solution in the form

z=A

4A=2

A=1/2

z_p(x)=1/2

y_p(x)=1/(2cos(x))

The general solution of the equation (1) has the form

y(x)=C₁sin(x)+C₂cos(2x)/cos(x)+1/(2cos(x))