1. ## Initial Value Problem

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

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

$\displaystyle y(0)=-1$

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

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

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

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

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

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

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

$\displaystyle y(0)=-1$

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

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

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

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

Then if I differentiation $\displaystyle y=\sin(x)\cos(x)-\cos(x)$ I obtain $\displaystyle \cos^2(x)-\sin^2(x)+\sin(x)$. 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 $\displaystyle y=\cos x\sin x-\sin x$ is a solution to $\displaystyle y'+y\tan x=\cos^2x$ , with $\displaystyle y(0)=-1$ , and it's easy to check that it is so...

Tonio

3. 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 $\displaystyle y=\cos x\sin x-\sin x$ is a solution to $\displaystyle y'+y\tan x=\cos^2x$ , with $\displaystyle y(0)=-1$ , 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 $\displaystyle y=\cos(x)\sin(x)-\sin(x)$ into the differential equation and then solved for the dervivative. Then I differeniated $\displaystyle y=\cos(x)\sin(x)-\sin(x)$ to verify that it's a solution. I made a simple algebraic error in here:

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