# Second Order Differential Equation

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• Feb 26th 2013, 12:12 PM
quantoembryo
Second Order Differential Equation
Determine the values of $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ given that $\displaystyle y=ax^2+bx+c$ is a solution to $\displaystyle x^2(y'')^2+y^2=80x^4+72x^3+50x^2+24x+16$ given the values $\displaystyle a$,$\displaystyle b$ and $\displaystyle c$ are positive.

$\displaystyle y'=2ax+b$ and $\displaystyle y''=2a$ hence $\displaystyle a^2x^4+2abx^3+(2ac+b^s+4a^2)x^2+c^2=80x^4+72x^3+50 x^2+24x+16$

However, from these results, I am not sure exactly how to determine which values you would use. I figured $\displaystyle a=\sqrt{80}$ and $\displaystyle c=4$, however, you get more than one option for $\displaystyle b$. Any help is greatly appreciated!
• Feb 26th 2013, 04:57 PM
Prove It
Re: Second Order Differential Equation
Why do you believe there has to only be one value for b?
• Feb 26th 2013, 05:00 PM
quantoembryo
Re: Second Order Differential Equation
Well, the answer depends on only one value for B.
• Feb 27th 2013, 08:06 AM
quantoembryo
Re: Second Order Differential Equation
I am supposed to give the result of sin(A)+sin(2B)+sin(3C); if I have more than one answer for B, that proves to be a bit difficult.