Constant Coefficient Differential Equation

[fa2uk]

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.

 

[jackmell]

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.

 

[LCKurtz]

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.

 

[fa2uk]

Thanks jackmell and LCKurtz

I changed the independent variable to x=e^t. 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.

 

[LCKurtz]

Thanks jackmell and LCKurtz

I changed the independent variable to x=e^t. 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 = e^t 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 e^2t 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 = y_c + y_p, all expressed in terms of t. Then substitute back for x to get the general solution and get your missing x²/30 term.

 

[fa2uk]

Hi LCKurtz

got it. thanks for the help and your time.