Constant Coefficient Differential Equation

fa2uk
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Hey there

I am new to Physics forums and could use some help understanding/solving this problem.

Use transformation x=e^t to convert equation
x^2y'' + 10xy' + 8y = x^2

Solve this equation to show that solution is

y = a/x^8 + b/x + x^2/30

Let me know if i missed anything here.
 
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It's not constant coefficient but you can transform it to a constant coefficient by your change of independent variable x=e^t. This is also an Euler-type equation, the homogeneous part is more easily solved by letting y=x^k, substituting, then figuring out what k is. That gives you the y_c=a/x+bx^{-8} part. However I assume you need to do the derivative substitutions where:

\frac{dy}{dx}=e^{-t}\frac{dy}{dt}

\frac{d^2y}{dx^2}=e^{-2t}\left(\frac{d^2 y}{dt^2}-\frac{dy}{dt}\right)

You can obtain those right? Then just substitute into the equation to obtain the constant-coefficient equation in the variable t, then let t=ln(x) to get it back into x


Then you need to solve the nonhomogeneous part which I guess reduction of order is the easiest approach.
 
jackmell said:
It's not constant coefficient but you can transform it to a constant coefficient by your change of independent variable x=e^t. This is also an Euler-type equation, the homogeneous part is more easily solved by letting y=x^k, substituting, then figuring out what k is. That gives you the y_c=a/x+bx^{-8} part. However I assume you need to do the derivative substitutions where:

\frac{dy}{dx}=e^{-t}\frac{dy}{dt}

\frac{d^2y}{dx^2}=e^{-2t}\left(\frac{d^2 y}{dt^2}-\frac{dy}{dt}\right)

You can obtain those right? Then just substitute into the equation to obtain the constant-coefficient equation in the variable t, then let t=ln(x) to get it back into x


Then you need to solve the nonhomogeneous part which I guess reduction of order is the easiest approach.

It would probably be easier to solve the NH equation while it is in terms of t using undetermined coefficients and transfer the whole thing back to x for the general solution.
 
Thanks jackmell and LCKurtz

I changed the independent variable to x=et. Yes, jackmell either solution works for the homogenous part.

However, I am stuck with x2/30 which I believe is the particular solution. i.e. I had set G(x) = x2 where the partcular solution is of the form Ax2+Bx+C.

jackmell, LCKurtz -> please could you elaborate your answers further.

Thanks for taking the time to respond.
 
fa2uk said:
Thanks jackmell and LCKurtz

I changed the independent variable to x=et. Yes, jackmell either solution works for the homogenous part.

However, I am stuck with x2/30 which I believe is the particular solution. i.e. I had set G(x) = x2 where the partcular solution is of the form Ax2+Bx+C.

jackmell, LCKurtz -> please could you elaborate your answers further.

Thanks for taking the time to respond.

After you made the x = et substitution, presumably you got the new DE:

\frac{d^2y}{dt^2}+ 9\frac{dy}{dt} + 8y = e^{2t}

which has complementary solution

y_c=Ae^{-8t}+Be^{-t}

Use undetermined coefficients to find a particular solution of this NH equation in t. Given you have e2t on the right you would look for a particular solution of the form

y_p = Ce^{2t}

Figure out C and you have the general solution y = yc + yp, all expressed in terms of t. Then substitute back for x to get the general solution and get your missing x2/30 term.
 
Hi LCKurtz

got it. thanks for the help and your time.
 
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