How to solve $t^2y''(t)-2ty'(t)+2y(t)=t^4\cos(t)$?
Differential equation : $t²y''(t)-2ty'(t)+2y(t)=t^4\cos(t)$
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ordinary-differential-equations
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0$t$ and $t^2$ are solutions – 2017-01-19
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0I do not know how to find the homogeneous solution. How is the characteristic equation? – 2017-01-19
3 Answers
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For Euler equation take $t=e^x$, so \begin{eqnarray} t^2y''&=&\frac{d^2y}{dx^2}-\frac{dy}{dx}\\ ty'&=&\frac{dy}{dx}\\ \end{eqnarray}
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Outline: This is an Euler equation. Try the a solution of the form $y = t^r$ in the homogeneous equation. After you plug it in, you can cancel all $t$'s and you have a quadratic in $r$. The two solutions for $r$ give you two solutions for the homogeneous equation. Then use variation of parameters to find a particular solution of the non-homogeneous equation.
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For the homogeneous, and trying solutions of the form $y=t^k,$ we get $y'=kt^{k-1}$ and $y''=k(k-1)t^{k-2}.$ So, $$t^2k(k-1)t^{k-2}-2tkt^{k-1}+2t^k=0$$
$$\Rightarrow k(k-1)-2k+2=0.$$ Now, find $k.$