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