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Tab'z's questions at Yahoo! Answers regarding a linear first order IVP

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Tab'z,

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.