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  1. #1
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    differential equation...

    4) Solve: \frac{dy}{dx} + y = x
    I don't even know what I'm supposed to do on this one...
    please help...I need this for my exam
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    Quote Originally Posted by jangalinn View Post
    4) Solve: \frac{dy}{dx} + y = x
    I don't even know what I'm supposed to do on this one...
    please help...I need this for my exam
    This equation is in the form y'+P(x)y = Q(x), thus we have to use the 'Integrating Factor' method.

    IF = e^{\int 1 \ \mathrm{d}x} = e^{x}

    Can you continue?

    Spoiler:
    1.) Multiply the original equation with the integrating factor.
    2.) Solve both sides of the equation. (By Integration).

    This website has a pdf file which explains the Integrating Factor method. However, it also has the solution to your question. It also has many other question which you could practice. Link
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  3. #3
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    Quote Originally Posted by jangalinn View Post
    4) Solve: \frac{dy}{dx} + y = x
    I don't even know what I'm supposed to do on this one...
    please help...I need this for my exam
    Since you are given an equation, I imagine you are supposed to find the general solution!

    There are two ways to do this:
    1) Find an integrating factor. That is, find some function u(x) so that multiplying the equation by u(x) makes the left side an "exact differential"- so that u(x)\frac{dy}{dx}+ u(x)y= \frac{d(u(x)y)}{dx}. Using the product rule on the right side of that, \frac{d(u(x)y)}{dx}= u(x)\frac{dy}{dx}+ \left(\frac{du}{dx}\right)y so our requirement on u(x) becomes u(x)\frac{dy}{dx}+ u(x)y= u(x)\frac{dy}{dx}+ \left(\frac{du}{dx}\right)y. That means we must have \frac{du}{dx}y= u(x)y or  \frac{du}{dx}= u, a separable equation for u. It "separates" as \frac{du}{u}= dx which integrates to ln(u)= x+ C. Taking exponentials of both sides, u(x)= C'e^x. We don't need the general solution so take C'= 1, u(x)= e^x.

    Now e^x (\frac{dy}{dx}+ y)= \frac{d(e^xy)}{dx}= xe^x
    integrating the left sides just gives e^xy while we can integrate the right side using "integration by parts": Let u= x, dv= e^x so that du= dx and v= e^x. \int xe^x dx= xe^x- \int e^xdx= xe^x- e^x= e^x(x- 1+ C. So e^xy= e^x(x-1)+ C and y= x- 1+ Ce^{-x}

    2) Separate into "homogeneous" and "non-homogeneous" parts. That is, first find the general solution to the "homogeneous" part, \frac{dy}{dx}+ y= 0. That is the same as \frac{dy}{dx}= -y which separates to \frac{dy}{y}= -dx. Integrating both sides, ln(y)= -x+ C so y= C' e^{-x} where C'= e^C.

    Now try to find a "specific", single, solution to \frac{dy}{dx}+ y= x. Since we are only trying to find a single solution rather than the general solution we can try "guessing". (educated guessing, of course!) Since the right hand side of the equation is "x", try y= Ax+ B. Then \frac{dy}{dx}= A and the equation becomes A+ Ax+ B= x. Since that must be true for all x, taking x= 0, we get A+ B= 0 or B= -A. Replacing B by -A gives Ax= x and, taking x= 1, A= 1 so B= -1. y(x)= x- 1 satisfies the differential equation \frac{dy}{dx}+ y= x

    Finally, we add the general solution to the homogenous equation to the specific solution to the non-homogeneous equation to get the general solution to the entire equation: y(x)= C'e^{-x}+ x- 1 as before.

    The second method is easier but requires more theory.
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