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June 2nd, 2017, 04:56 PM   #1
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Linear Differential equation with constant coefficient

A solution of $(D^2+1)^3y=\sin x$ is
A) $\displaystyle \sum_{n=1}^{3} (c_nx^{n-1}\sin x+d_nx^{n-1}\cos x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$.
B) $\displaystyle \sum_{n=1}^{3} (c_n \sin^n x+d_n \cos^n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$.
C) $\displaystyle \sum_{n=1}^{3} (c_n \sin n x+d_n \cos n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$.
D) None of the above.

My answer is Option B. Please check and let me know if wrong.

Thanks.

Last edited by skipjack; June 2nd, 2017 at 11:39 PM.
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June 3rd, 2017, 03:32 AM   #2
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Quote:
Originally Posted by Lalitha183 View Post
A solution of $(D^2+1)^3y=\sin x$ is
A) $\displaystyle \sum_{n=1}^{3} (c_nx^{n-1}\sin x+d_nx^{n-1}\cos x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$.
B) $\displaystyle \sum_{n=1}^{3} (c_n \sin^n x+d_n \cos^n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$.
C) $\displaystyle \sum_{n=1}^{3} (c_n \sin n x+d_n \cos n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$.
D) None of the above.

My answer is Option B. Please check and let me know if wrong.

Thanks.
Please someone help!!
Tomorrow I have an exam
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June 4th, 2017, 12:28 PM   #3
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Are you really required to use a "series solution" method? I would use that to try to solve a linear differential equation with variable coefficients, but this, $\displaystyle (D^2+ 1)^3y= \sin(x)$, has constant coefficients. The "characteristic equation" for the associated homogeneous equation ($\displaystyle (D^2+1)^3= 0$) is $\displaystyle (r^2+ 1)^3= 0$. $r= i$ is a triple root and $r= -i$ is another triple root. That means that the general solution to the associated homogeneous equation is $\displaystyle y(x)= A \cos(x)+ B \sin(x)+ C x \cos(x)+ D x \sin(x)+ E x^2 \cos(x)+ F x^2 \sin(x)$.

The "non-homogeneous" part of the equation is $\displaystyle \sin(x)$ which is already a solution to the homogeneous equation, as well as $x$ and $\displaystyle x^2$ times $\displaystyle \sin(x)$. So try $\displaystyle y(x)= Gx^3 \cos(x)+ Hx^3 \sin(x)$.

The calculations from this point are tedious but straightforward.

Last edited by skipjack; June 5th, 2017 at 12:36 AM.
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