# DIFFERENTIAL EQUATIONS -Can you help me ?

• Feb 9th 2009, 06:22 AM
butbi9x
DIFFERENTIAL EQUATIONS -Can you help me ?
Verify that y = sinxcosx - cosx is a solution of the initial-value problem y' + (tanx)y= cos^2x , y(0) = -1 on the interval - - П/2 < x< П/2
• Feb 9th 2009, 06:37 AM
Bilbo Baggins
Quote:

Originally Posted by butbi9x
Verify that y = sinxcosx - cosx is a solution of the initial-value problem y' + (tanx)y= cos^2x , y(0) = -1 on the interval - - П/2 < x< П/2

Hi. I'm new to differential equations; however, I think I see the solution. Check it out:

The idea here is to try and make one side look like the derivative of a product. Take the left side and multiply by the function sec X. Carrying that over the entire equation yields: SecX*y' + secXtanX*y =cosX

This is clearly the deritive of the product secX*y. Simply integrate both sides and get the new equation: secX*y=sinX +c. Just divide by secX and you receive: Y=sinXcosX+cosX
• Feb 9th 2009, 03:21 PM
Jake8054
Use method of linear equations

find integrating factor first: IF=exp(integral(p(x)))

exp(integral(tanx))=secx

y=cosx(integral(cosx))

y=sinxcosx+kcosx

use initial value: -1=k

y=sinxcosx-cosx or...

y=(1/2)sin2x -cosx