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This is the final part of a question: c)Find the solution of the differential equation $$x\cdot y''+2(1+2x)y'+4(1+x)y=32\exp(2x)$$ for which y=2exp(2) and y'=0 at x=1.

Here is the rest of the question:

a)Show that the substitution v=xy transforms the differential equation $$x\cdot y''+2(1+2x)y'+4(1+x)y=32\exp(2x) $$ $$x≠0$$ into the differential equation $$v''+4v'+4v=32\exp(2x)$$

b)Given that $v=a\exp(2x)$, where $a$ is a constant, is a particular integral of this transformed equation, find $a$. $(a=2)$

I know how to do the first two part but don't know how to continue.

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If you have found $v$, both homogeneous and particular, and know that $v = xy$ ... –  Eugene Shvarts Jul 20 '12 at 8:00
    
My guess would be solve the equation for v, then y=v/x. –  Mike Jul 20 '12 at 8:22
    
Could you make it an answer or be more specific.? –  Vic. Jul 20 '12 at 8:36
    
You said you were able to show that the substitution $v=xy$ transforms the equation into $v''+4v'+4v=32e^{2x}$, correct? I assume you already know how to solve an equation like this. –  Mike Jul 20 '12 at 8:47
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2 Answers 2

up vote 2 down vote accepted

Taking over from the comment on Babak Sorouh's answer. So you say you can get

$$v=C_1e^{-2x}+C_2xe^{-2x}+2e^{2x}$$

Now let's solve for y.

$$xy=C_1e^{-2x}+C_2xe^{-2x}+2e^{2x}$$

$$y=C_1x^{-1}e^{-2x}+C_2e^{-2x}+2x^{-1}e^{2x}$$

Next we need a formula for $y'$.

$$y'=C_1(-x^{-2}e^{-2x}-2x^{-1}e^{-2x})-2C_2e^{-2x}+4x^{-1}e^{2x}-2x^{-2}e^{2x}$$

Now we take our formulas and plug in the given values

$$C_1e^{-2}+C_2e^{-2}+2e^2=2e^2$$

$$C_1+C_2=0,C_1=-C_2$$

$$C_1(-e^{-2}-2e^{-2})-2C_2e^{-2}+4e^2-2e^2=0$$

$$-3C_1e^{-2}-2C_2e^{-2}=-2e^2$$

$$3C_1+2C_2=2e^4$$

$$3C_1-2C_1=2e^4$$

$$C_1=2e^4,C_2=-C_1=-2e^4$$

So this gives us a final answer of

$$y=2e^4x^{-1}e^{-2x}-2e^4e^{-2x}+2x^{-1}e^{2x}=$$

$$\frac{2e^{4-2x}+2e^{2x}}{x}-2e^{4-2x}$$

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Thanks for finalizing my answer nicely. –  B. S. Jul 20 '12 at 10:44
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If it is indicated you take $v=xy$ so we have $$v'=y+xy'\longrightarrow v''=2y'+xy''$$ Now putting them into the equation $x\cdot y''+2(1+2x)y'+4(1+x)y=32 e^{2x} $ we have $$v''-2\frac{v'-y}{x}+2(1+2x)\frac{v'-y}{x}+4(1+x)y=32e^{2x}$$ Simplifing gives $$v''+4v'+4v=32e^{2x}, x\neq0$$ Now asuume that $v=ae^{2x}$ for some $a$ so you get $$v'=2ae^{2x}\longrightarrow v''=4ae^{2x}$$ and if you put the results into the converted equation, you have $$16ae^{2x}=32e^{2x}$$ This means that $a=2$. Now you have this equation $$v''+4v'+4v=32\exp(2x)$$ The equation is a linear equation with constant coefficients of second order. It is not homogenous so you have to find a general solution for the homogenous one first and a particular solution as well. Here, in the problem the particular solution is given so, we just regard the equation $$v''+4v'+4v=0$$. By a suitable auxiliary equation you will find $v_c(x)=C_1e^{-2x}+C_2xe^{-2x}$ and so the general solution of the nonhomogeneous equation on any interval with $x\neq 0$ is $$v(x)=v_c(x)+v_p(x)=C_1e^{-2x}+C_2xe^{-2x}+2e^{2x}$$ I think you can satisfy the initial conditions. Note that $y=\frac{v(x)}{x}$.

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I know how to get $v(x)=v_c(x)+v_p(x)=C_1e^{-2x}+C_2xe^{-2x}+2e^{2x}$ The problem is I don't know how to use "$y=2exp(2)$ and $y'=0$ at $x=1$." to get the value of C1 and C2. –  Vic. Jul 20 '12 at 9:29
    
Good answer! :+) –  amWhy Mar 12 '13 at 1:50
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