Finding y using complimentary and particular solutions

[John777]

Homework Statement

y''+4y = 2x^2+2sinx Find Y.

Homework Equations

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The Attempt at a Solution

Well I know the complimentary solution is Y_c=C₁e^2ix+C₂e^-2ix

I am just very confused on finding the y_p. I know the basic method is find y_p then take second derivative and plug in, but I'm not sure what the initial equation is for y_p thus I cannot proceed. If you answer this please don't just write the initial equation for y_p. Please let me know how you got it. Thanks you guys are a lifesavers!

edit: I believe the form of y_p is something like Ax^2+B+Dcosx-Esinx As this is just from another example I had, but I do not know how they got this.

 

[D H]

Have you tried to get lucky?

The lucky guess method works quite nicely here.

 

[John777]

Have you tried to get lucky?

The lucky guess method works quite nicely here.

I've never head of that method, can you please ellaborate?

 

[D H]

Google is your friend. It's also known as the method of undetermined coefficients.

 

[HallsofIvy]

edit: I believe the form of yp is something like Ax^2+B+Dcosx-Esinx As this is just from another example I had, but I do not know how they got this.

Almost. For "x^2" you need to try something like Ax^2+ Bx+ C. That is, the power of x and all lower powers. So try y= Ax^2+ Bx+ C+ Dcos(x)+ Esin(x). Plug that into the differential equation and solve for A, B, C, D, and E.

(They "got that" by knowing that the derivative of a power of x involves a lower power of x ((x^n)'= n x^(n-1)) and that the derivative of sine is a cosine and vice-versa.)

 

[John777]

Almost. For "x^2" you need to try something like Ax^2+ Bx+ C. That is, the power of x and all lower powers. So try y= Ax^2+ Bx+ C+ Dcos(x)+ Esin(x). Plug that into the differential equation and solve for A, B, C, D, and E.

(They "got that" by knowing that the derivative of a power of x involves a lower power of x ((x^n)'= n x^(n-1)) and that the derivative of sine is a cosine and vice-versa.)

OK I get it and have solved for the A, B, C, D, and E variables. Is it possible to solve for the C1 and C2 into the complimentary solution?

 

[D H]

Only if you are told about some specific values of the function.

Consider y'-y=0, which has y=C*e^x as a solution. Any constant value for C will yield a solution to the differential equation. You can solve for C if you are told the value of the function for some particular x.

 

[John777]

Only if you are told about some specific values of the function.

Consider y'-y=0, which has y=C*e^x as a solution. Any constant value for C will yield a solution to the differential equation. You can solve for C if you are told the value of the function for some particular x.

Thanks for all the help. I really appreciate it.

One more question, if I had an equation such as 3"+7y'-6y = 6e^-4x

What would the lucky guess method initial y particular be?

Also what if the left side was x6e^-4x

 

[HallsofIvy]

I assume you mean y"+ 7y'+ 6y. That gives a characteristic equation of $r^2+ 7r+ 6= (r+ 6)(r+ 1)= 0$ and so two independent solutions to the associated homogeneous equation are $e^{-6x}$ and $e^{-x}$. Since the right hand side does not involve either of those, try a particular solution of the form $y= Ae^{-4x}$.

For $6xe^{-4x}$, as I said before, use x and lower powers: try $y= (Ax+ B)e^{-4x}$.