# 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

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Troy Lesure
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
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
$$\displaystyle{y}={y}_{{h}}+{y}_{{{p}{1}}}+{y}_{{{p}{2}}}+\ldots$$