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

AI Thread Summary
The discussion centers on solving the initial value problem (IVP) for the first-order linear ordinary differential equation (ODE) given by y' + tan(x)y = cos²(x) with the initial condition y(0) = 2. The integrating factor is calculated as μ(x) = sec(x), which transforms the ODE into a form that allows for straightforward integration. After integrating, the solution is expressed as y(x) = cos(x)(sin(x) + 2), determined by applying the initial condition to find the constant C. The thread encourages further engagement by inviting users to post additional differential equations problems for discussion. The solution effectively demonstrates the method for tackling similar first-order linear ODEs.
MarkFL
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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.
 
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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.
 
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