Infinite Series Method 2nd order DE

The following second order Differential Equations must be solved with the appropreate Infinite Series Method. You may verify DE with other method only after work is shown step by step using the infinite series methods. Scanned work is ok as long as I can read it. These problems will require a lot of steps so I added a couple of ...continues

Solving differential equations with the Laplace Transform

Use Laplace Transforms to solve problems below. Please show all work step by step I am using your work as a study guide for my upcoming Final, so please explain well. Scannned work is ok as long as I can read it. Use Laplace Transforms to solve DE's. 1) y" - 8y' + 20y = t(e^t) , y(0)=0 , y'(0)=0 2) y''' + 2y ...continues

Modeling with higher order Differential Equations

(See attached file for full problem description)

Laplace Transform

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Linear differential operator

Problem states " if L[y] =ay'' +by' +cy, where a,b,and c are constants, compute L[e^rx], where "r" is constant. Is this just a matter of substituting for "y"? Please work out, thanks!

Substituting variable

Please show how to solve y’’ – 3y^2=0, substituting v=y’ so y’’ = v dv/dy Initial conditions are y(0) =2 and y’(0)=4 I got it as far as dy/dx = (y^3 +c)^1/2 but that might be wrong!

Laplace Transforms

Please show work by step by step. Scanned work is OK as long as I can read it.

Laplace Transforms

PLease show all work Step by Step. Scanned work is OK as long as I can read it. Two problems attached please complete both.

Practice Problems for Exam

These are problems from the text that were advised to study for the next exam. I would appreciate that all the work be shown as well as the answers so that I can follow the problem. Thanks

Substitution of "v" in a 2nd order d.e.

I have been tasked with solving y’’ – 3y^2 =0 using the technique used substituting v for y’, therefore substituting v dv/dy for y’’. (Equation with “x” missing) I broke it down as follows Y’’ -3y^2 =0 Y’’ =3y^2 Substituting I get v dv/dy = 3y^2 Separating variables, I get v dv =3y^2dy Integrating I get 1/2v^2 +c = y^3 +c ...continues