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Solve for $2y''+3y'+y=t$ using method of undetermined coefficients. So I let $Y=At+b$ to solve for the particular solution. After substituting the first and second derivative into $2y''+3y'+y=t$, I get $3A+At+B= t,$ -> $ A(3+t)+B=t$. But I don't know how to proceed from here because computing $3+t=1$ gives $t=-2$ whereas the solution is $Y=t-3$. Can someone help correct my mistake?

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  • $\begingroup$ Why are you taking $3+t=1$? $\endgroup$ – Git Gud Apr 13 '14 at 20:22
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I believe the mistake you're suspecting was when you said you got $3A + At+B=t$. When we let $y=At+B$ (which also means $y'=A$ and $y''=0$), we should have \begin{align} 2y''+3y'+y=t &\implies 2(0)+3A+(At+B)=t\\ &\implies At+(3A+B)=t \end{align} When we equate coefficients, I got $A=1$ (from $At=t$) and $B=-3$ (from $3A+B=0$).

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  • $\begingroup$ For $y=At+B$, wouldn't we have $y'=t$ and $y''=0$ or am I missing something? $\endgroup$ – user67527 Apr 13 '14 at 20:31
  • $\begingroup$ You're right; I made a mistake. But I edited and hope I fixed my answer. $\endgroup$ – Cookie Apr 13 '14 at 20:40

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