# math

Solve the following ordinary differential equations.
I) (2y+x^2+1)dy/dx+2xy-9x^2=0
ii) d^2/dx^2+3day/DX+2y=x^2

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1. #1:
(2y+x^2+1)dy/dx + (2xy-9x^2) = 0
We can rewrite this as
(2y + x^2 + 1)dy + (2xy - 9x^2)dx = 0
Check for exactness:
Let M = 2xy - 9x^2
Let N = 2y + x^2 + 1
∂M / ∂y = ∂N / ∂x
∂(2xy - 9x^2) / ∂y = ∂(2y + x^2 + 1) / ∂x
2x = 2x
Thus it is indeed exact. To further solve,
∂F/∂x = M = 2xy - 9x^2
∫ ∂F = ∫ (2xy - 9x^2)∂x
F = (x^2)y - 3x^3 + g(y)

To get g(y), we differentiate it partially with respect to y:
∂F/∂y = ∂/∂y((x^2)y - 3x^3 + g(y)) = N
∂F/∂y = x^2 + g'(y) = N
x^2 + g'(y) = 2y + x^2 + 1
g'(y) = 2y + 1
Integrating,
g(y) = y^2 + y + C
Therefore,
F = (x^2)y - 3x^3 + y^2 + y + C

#2: This is incomplete. I can't remember solving the right side (if there is a function of x; in this case it's the x^2). I have to browse my college notes first lol:
d^2y/dx^2 + 3(dy/dx) + 2y = x^2
Rewriting, such that dy/dx = D:
(D^2 + 3D + 2)y = x^2
Left side becomes:
m^2 + 3m + 2
Factoring,
(m+1)(m+2)
m = -1 and m = -2
The general solution for something like this is,
yc = c,1*e^(m,1*x) + c,2*e^(m,2*x) + ... + c,n*e^(m,n*x)
where m is the root.
Thus, the solution for the left side, (D^2 + 3D + 2)y, is
yc = c,1*e^(-x) + c,2*e^(-2x)

I hope someone checks my work also.
hope this helps? ;u;

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2. so far so good. The complete solution turns out to be, as I'm sure you have worked out:

y = c1 e^-x + c2 e^-2x + 1/4 (2x^2-6x+7)

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