y''(x)+4y'(x)+3y(x)=9x^2-20+30e^{2x},\ y(0)=0,\ y'(0)=2I

Jessica Scott 2021-11-23 Answered

\(\displaystyle{y}{''}{\left({x}\right)}+{4}{y}'{\left({x}\right)}+{3}{y}{\left({x}\right)}={9}{x}^{{2}}-{20}+{30}{e}^{{{2}{x}}}\),\(\ y(0)=0,\ y'(0)=2\)
I'm stuck on finding a particular integral for the non-homogeneous R.H.S.: \(\displaystyle{9}{x}^{{2}}-{20}+{30}{e}^{{{2}{x}}}\)
So \(\displaystyle{y}={y}_{{c}}+{y}_{{p}}\). I found \(\displaystyle{y}_{{c}}={C}_{{1}}{e}^{{-{x}}}+{C}_{{2}}{e}^{{-{3}{x}}}\) (where \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\) are constants) But what would I try for \(\displaystyle{y}_{{p}}\)?? If it helps, I know how to do it if it's just a polynomial (i.e. for a 2nd degree polynomial): you set \(\displaystyle{y}_{{p}}={A}{x}^{{2}}+{B}{x}+{C}\) and \(\displaystyle{y}'_{{p}}={A}{x}+{B}\) and \(\displaystyle{y}{''}_{{p}}={A}\)
But in this case we have different things. What would I need to do? I would really appreciate your help. Tha
in advance!

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Expert Answer

Troy Lesure
Answered 2021-11-24 Author has 1383 answers
Hint
Let
\(\displaystyle{y}_{{p}}={a}{x}^{{2}}+{b}{x}+{c}+{k}{e}^{{{2}{x}}}\)
substitute this in the equation and proceed through.
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Linda Tincher
Answered 2021-11-25 Author has 255 answers
Solving this
\(\displaystyle{y}{''}{\left({x}\right)}+{4}{y}'{\left({x}\right)}+{3}{y}{\left({x}\right)}={9}{x}^{{2}}-{20}+{30}{e}^{{{2}{x}}},\ {y}{\left({0}\right)}={0},\ {y}'{\left({0}\right)}={2}\)
Is the same as solving
\(\displaystyle{y}{''}{\left({x}\right)}+{4}{y}'{\left({x}\right)}+{3}{y}{\left({x}\right)}={9}{x}^{{2}}-{20}\)
using \(\displaystyle{y}_{{p}}={A}{x}^{{2}}+{B}{x}+{C}\) and
\(\displaystyle{y}{''}{\left({x}\right)}+{4}{y}'{\left({x}\right)}+{3}{y}{\left({x}\right)}={30}{e}^{{{2}{x}}}\)
using \(\displaystyle{y}_{{p}}={R}{e}^{{{2}{x}}}\)
The solution is just going to be the sum of all particular solutions
\(\displaystyle{y}={y}_{{h}}+{y}_{{{p}{1}}}+{y}_{{{p}{2}}}+\ldots\)
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