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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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    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.2012-07-20

2 Answers 2

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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.2012-07-20
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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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    Good answer! :+)2013-03-12