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Here is a link to the question:Can someone help me solve this:

y'+tan(x)y=cos^2(x), y(0)=2?

Can someone help me solve this: y'+tan

I have posted a link there to this topic so the OP can find my response.

- Thread starter MarkFL
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- Thread starter
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- #1

Here is a link to the question:Can someone help me solve this:

y'+tan(x)y=cos^2(x), y(0)=2?

Can someone help me solve this: y'+tan

I have posted a link there to this topic so the OP can find my response.

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We are given to solve the IVP:

\(\displaystyle \frac{dy}{dx}+\tan(x)y=\cos^2(x)\) where \(\displaystyle y(0)=2\)

We should observe that the ODE is a first order linear equation, and so we want to compute the integrating factor as follows:

\(\displaystyle \mu(x)=e^{\int\tan(x)\,dx}=\sec(x)\)

Multiplying the ODE by the integrating factor, we have:

\(\displaystyle \sec(x)\frac{dy}{dx}+\sec(x)\tan(x)y=\cos(x)\)

Now, recognizing that the left side is the differentiation of a product, we may write:

\(\displaystyle \frac{d}{dx}\left(\sec(x)y \right)=\cos(x)\)

Integrating with respect to $x$, we find:

\(\displaystyle \int\,d\left(\sec(x)y \right)=\int\cos(x)\,dx\)

\(\displaystyle \sec(x)y=\sin(x)+C\)

\(\displaystyle y(x)=\sin(x)\cos(x)+C\cos(x)\)

Now, using the given initial condition, we may find the parameter $C$:

\(\displaystyle y(0)=\sin(0)\cos(0)+C\cos(0)=C=2\)

Hence, the solution satisfying the given IVO is:

\(\displaystyle y(x)=\sin(x)\cos(x)+2\cos(x)=\cos(x)(\sin(x)+2)\)

To Tab'z and any other guests viewing this topic, I invite and encourage you to post other differential equations problems here in our Differential Equations forum.

Best Regards,

Mark.