Differential Equation

[uberathlete]

Hi everyone. I'm having trouble figuring out how to go about this problem:

Find all the solutions of the form y = ax^2 + bx + c to the DE
(y'(x))^2 + 4x = 3y(x) + x^2 + 1.

I tried bringing the 3y(x) over to the left side and the 4x to the right, but then I get stuck. I can't really square root it either. I tried dividing by y but again, I get stuck.

If anyone could help me out with this, it'd be greatly appreciated. Thanks!

 

[galactus]

After you bring the like terms onto their respective sides

\(\displaystyle (\frac{dy}{dx})^{2}-3y=x^{2}-4x+1\),

try using the integrating factor to solve.

I believe this is it:

\(\displaystyle e^{\int{-3}dx}=e^{-3x}\)

 

[Unco]

Find all the solutions of the form y = ax^2 + bx + c to the DE
(y'(x))^2 + 4x = 3y(x) + x^2 + 1.

I tried bringing the 3y(x) over to the left side and the 4x to the right, but then I get stuck. I can't really square root it either. I tried dividing by y but again, I get stuck.

Plug y=ax^2+bx+c and its derivative into the DE. Equate coefficients (of x^2, x and constant) to solve for a, and then use these for b, and these for c, respectively.

 

[uberathlete]

After you bring the like terms onto their respective sides

\(\displaystyle (\frac{dy}{dx})^{2}-3y=x^{2}-4x+1\),

try using the integrating factor to solve.

I believe this is it:

\(\displaystyle e^{\int{-3}dx}=e^{-3x}\)

Hi Galactus. I was sort of thinking the same thing, but the dy/dx term is squared. Wouldn't that make the equation non-linear and so I wouldn't be able to use the integrating factor?

 

[uberathlete]

Find all the solutions of the form y = ax^2 + bx + c to the DE
(y'(x))^2 + 4x = 3y(x) + x^2 + 1.

I tried bringing the 3y(x) over to the left side and the 4x to the right, but then I get stuck. I can't really square root it either. I tried dividing by y but again, I get stuck.

Plug y=ax^2+bx+c and its derivative into the DE. Equate coefficients (of x^2, x and constant) to solve for a, and then use these for b, and these for c, respectively.

Hmm ... interesting. Lemme try it!

 

[galactus]

After you bring the like terms onto their respective sides

\(\displaystyle (\frac{dy}{dx})^{2}-3y=x^{2}-4x+1\),

try using the integrating factor to solve.

I believe this is it:

\(\displaystyle e^{\int{-3}dx}=e^{-3x}\)

Hi Galactus. I was sort of thinking the same thing, but the dy/dx term is squared. Wouldn't that make the equation non-linear and so I wouldn't be able to use the integrating factor?

Yes, you're correct. An oversight on my part. The heck with all those fancy-schmancy DE techniques. Unco seen it. They gave you the answer...\(\displaystyle ax^{2}+bx+c\).
Unco's method is the way to go with this problem. I ran through the calculations and it worked great. Once you have your answers, sub back into the original and check.