I'm trying to find the general solution to: y''+4y=t^2+7e^t

Clifton Sanchez 2021-11-20 Answered
I'm trying to find the general solution to:
\(\displaystyle{y}{''}+{4}{y}={t}^{{2}}+{7}{e}^{{t}}\)
The actual problem wants me to find the initial value problem with y(0) = 0 and y'(0) = 2 but I'm confident that I can find the IVP after finding the general solution.
What I DO need help with is this:
I'm trying to set \(\displaystyle{Y}{\left({t}\right)}={A}{t}^{{2}}+{B}{t}+{C}\) and solving for A, B, and C for a specific solution but I find two different values for A. (1/4, and 0).
I did solve for \(\displaystyle{e}^{{t}}\) and found the answer to be 7/5.

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Expert Answer

Charles Clute
Answered 2021-11-21 Author has 463 answers
the characteristic equation is \(\displaystyle{m}^{{2}}+{4}={0}\ {m}=\pm{2}{i}\)
\(\displaystyle{y}_{{c}}={C}_{{1}}{\cos{{2}}}{t}+{C}_{{2}}{\sin{{2}}}{t}\)
\(\displaystyle{y}_{{p}}={A}{t}^{{2}}+{B}{t}+{C}+{D}{e}^{{t}}\)
\(\displaystyle{y}'_{{p}}={2}{A}{t}+{B}+{D}{e}^{{t}}\)
\(\displaystyle{y}{''}_{{p}}={2}{A}+{D}{e}^{{t}}\)
substitute in the original equation
\(\displaystyle{2}{A}+{D}{e}^{{t}}+{4}{A}{t}^{{2}}+{4}{B}{t}+{4}{C}+{4}{D}{e}^{{t}}={t}^{{2}}+{7}{e}^{{t}}\)
so \(\displaystyle{A}=\frac{{1}}{{4}}\)
\(\displaystyle{B}={0}\)
\(\displaystyle{C}=-\frac{{1}}{{8}}\)
\(\displaystyle{D}=\frac{{7}}{{5}}\)
the general solution is
\(\displaystyle{y}={C}_{{1}}{\cos{{2}}}{t}+{C}_{{2}}{\sin{{2}}}{t}+{\frac{{{t}^{{2}}}}{{{4}}}}-{\frac{{{1}}}{{{8}}}}+{\frac{{{7}}}{{{5}}}}{e}^{{t}}\)
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Florence Pittman
Answered 2021-11-22 Author has 1186 answers
The characteristic equation is \(\displaystyle{r}^{{2}}+{4}={0}\), so you get \(\displaystyle{y}_{{h}}={c}_{{1}}{\cos{{\left({2}{t}\right)}}}+{c}_{{2}}{\sin{{\left({2}{t}\right)}}}\). Now, use the method of undetermined coefficients to make some guesses to the general solution: If you plug in \(\displaystyle{y}_{{3}}={c}_{{3}}{e}^{{t}}\), you'll get \(\displaystyle{y}{''}_{{3}}+{4}{y}_{{3}}={5}{c}_{{3}}{e}^{{t}}\) , so having \(\displaystyle{c}_{{3}}={\frac{{{7}}}{{{5}}}}\) gives the \(\displaystyle{e}^{{t}}\) on the right. Now, for the \(\displaystyle{t}^{{2}}\), since you have a quadratic, you should guess \(\displaystyle{y}{''}_{{4}}+{4}{y}^{{4}}={a}{t}^{{2}}\). Then, \(\displaystyle{y}{''}_{{4}}+{4}{y}_{{4}}={2}{a}+{4}{\left({a}{t}^{{2}}+{b}{t}+{c}\right)}={4}{a}{t}^{{2}}+{b}{t}+{c}+{2}{a}\). Matching this to \(\displaystyle{t}^{{2}}\) gives \(\displaystyle{a}={\frac{{{1}}}{{{4}}}},\ {b}={0},\ {c}=-\frac{{1}}{{8}}\)
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