Solving IVP for Second-Order Differential Equation: x^2*y+4*x*y'+4*y=0

[BobMarly]

Not sure of solution for IVP

Homework Statement

(x^2)*y"+4*x*y'+4*y=0
y(1)=1 y'(1)=2

Homework Equations

Start with r(r-1)+4r+4=0
then (r^2)+3r+4=0
get (-3+/-7i)/2

The Attempt at a Solution

Leads to y=C(1)e^(-3t/2)cos(7t/2)+C(2)e^(-3t/2)sin(7t/2)
More y'=... one large formula
Am I on the correct track, is there something I missed, or is there a shortcut here
it seems that by putting in IV's, it doesn't lead to anything clean

 

[Mark44]

Homework Statement

(x^2)*y"+4*x*y'+4*y=0
y(1)=1 y'(1)=2

Homework Equations

Start with r(r-1)+4r+4=0
then (r^2)+3r+4=0
get (-3+/-7i)/2

The Attempt at a Solution

Leads to y=C(1)e^(-3t/2)cos(7t/2)+C(2)e^(-3t/2)sin(7t/2)

First off, does this function satisfy your differential equation? If so, you have the right general solution.

Next evaluate your solution and y'(t) at 1 and solve for the two constants. The check there is verifying that y(1) = 1 and y'(1) = 2.

At this stage in your learning, you should get in the habit of verifying that solutions you find are actually solutions. You've already done all the hard work, so checking your work is simply a matter of 1) showing that your solution satisfies the DE, and 2) that your solution and its derivative satisfies the initial conditions.

More y'=... one large formula
Am I on the correct track, is there something I missed, or is there a shortcut here
it seems that by putting in IV's, it doesn't lead to anything clean

 

[BobMarly]

How do I know it satisfies the equation?

 

[LCKurtz]

Homework Statement

(x^2)*y"+4*x*y'+4*y=0
y(1)=1 y'(1)=2

Homework Equations

Start with r(r-1)+4r+4=0
then (r^2)+3r+4=0
get (-3+/-7i)/2

The Attempt at a Solution

Leads to y=C(1)e^(-3t/2)cos(7t/2)+C(2)e^(-3t/2)sin(7t/2)

No it doesn't.

More y'=... one large formula
Am I on the correct track, is there something I missed, or is there a shortcut here
it seems that by putting in IV's, it doesn't lead to anything clean

You are confusing the characteristic equation for constant coefficient DE's with this problem. The "solution" you propose is what you would get for those roots of r if this were a constant coefficient equation. But it isn't. This is an Euler equation and you get that indicial equation by looking for a solution $y = x^r$. Look in your text for the form the solution takes for complex conjugate roots. It will have x's times cosines and sines of ln x's.

 

[HallsofIvy]

In fact, the reason the "Cauchy-Euler" equation is as simple as the equation with constant coefficients (and easily confused with it) is that the substitution $t= e^x$ converts the Cauchy-Euler equation $Ax^2d^2y/dx^2+ Bx dy/dx+ Cy= 0$ into the 'constant coefficients' equation [itex]Ad^2y/dt^2+ (B- A)dy/dt+ Cy= 0.
The two have the same characteristic equation. In your case the constant-coefficients solution is
$$e^{-3t/2}(C_1cos(7t/2)+ C_2sin(7t/2))$$
so that the solution to the original CauchyEuler equation is
$$e^{-3(ln(x))/2}(C_1cos(7ln(x)/3)+ C_2 sin(7ln(x)/3))= x^{-3/2}(C_1cos(7ln(x)/2)+ C_2sin(7ln(x)/2))$$