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

[MarkFL]

Here is the question:

Can someone help me solve this:

y'+tan⁡(x)y=cos^2⁡(x), y(0)=2?

Here is a link to the question:

Can someone help me solve this: y'+tan

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

 

[MarkFL]

Hello Tab'z,

We are given to solve the IVP:

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

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

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

Multiplying the ODE by the integrating factor, we have:

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

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

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

Integrating with respect to $x$, we find:

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

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

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

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

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

Hence, the solution satisfying the given IVO is:

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

To Tab'z and any other guests viewing this topic, I invite and encourage you to post other differential equations problems here in our http://www.mathhelpboards.com/f17/ forum.

Best Regards,

Mark.