Initial Value Problem

**adkinsjr** · December 20th 2009, 10:08 PM

Verify that $y=\sin(x)\cos(x)-\cos(x)$ is solution of the initial value problem:

$y'+y\tan(x)=\cos^2(x)$

$y(0)=-1$

$-\frac{\pi}{2}<x<\frac{\pi}{2}$
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I don't understand the object of this problem. Obviously the differential equation gives the derivative:

$y'=\cos^2(x)-y\tan(x)$

$=\cos^2(x)-\left(\sin(x)\cos(x)-\cos(x)\right)\tan(x)$

$=\cos^2(x)-\sin^2(x)-\sin(x)$

Then if I differentiation $y=\sin(x)\cos(x)-\cos(x)$ I obtain $\cos^2(x)-\sin^2(x)+\sin(x)$ . This doesn't agree with the other equation, although it is close.

**tonio** · December 21st 2009, 02:21 AM

Originally Posted by adkinsjr

Verify that $y=\sin(x)\cos(x)-\cos(x)$ is solution of the initial value problem:

$y'+y\tan(x)=\cos^2(x)$

$y(0)=-1$

$-\frac{\pi}{2}<x<\frac{\pi}{2}$
--------------------------------------------------------------------------------------------------
I don't understand the object of this problem. Obviously the differential equation gives the derivative:

$y'=\cos^2(x)-y\tan(x)$

$=\cos^2(x)-\left(\sin(x)\cos(x)-\cos(x)\right)\tan(x)$

$=\cos^2(x)-\sin^2(x)-\sin(x)$

Then if I differentiation $y=\sin(x)\cos(x)-\cos(x)$ I obtain $\cos^2(x)-\sin^2(x)+\sin(x)$ . This doesn't agree with the other equation, although it is close.

I don't understand what you did...apparently you solved the differential equation, but you didn't have to do that. You had to check that $y=\cos x\sin x-\sin x$ is a solution to $y'+y\tan x=\cos^2x$ , with $y(0)=-1$ , and it's easy to check that it is so...

Tonio

**adkinsjr** · December 21st 2009, 02:49 AM

Originally Posted by tonio

I don't understand what you did...apparently you solved the differential equation, but you didn't have to do that. You had to check that $y=\cos x\sin x-\sin x$ is a solution to $y'+y\tan x=\cos^2x$ , with $y(0)=-1$ , and it's easy to check that it is so...

The second term is -sin(x)
Tonio

I see what I did wrong. I substituted $y=\cos(x)\sin(x)-\sin(x)$ into the differential equation and then solved for the dervivative. Then I differeniated $y=\cos(x)\sin(x)-\sin(x)$ to verify that it's a solution. I made a simple algebraic error in here:

This should be $\cos^2(x)-\sin^2(x)+\sin(x)$ .

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