Question

Solve the differential equation. (x and y are variables) (D^2+4D)y=1+2x+3x^2

Multivariable functions
ANSWERED
asked 2021-09-14
Solve the differential equation. (x and y are variables)
\(\displaystyle{\left({D}^{{2}}+{4}{D}\right)}{y}={1}+{2}{x}+{3}{x}^{{2}}\)

Expert Answers (1)

2021-09-15

Given differential equation is \(\displaystyle{\left({D}^{{2}}+{4}{D}\right)}{y}={1}+{2}{x}+{3}{x}^{{2}}\) - (1)
\(\displaystyle{y}\text{}{4}{y}'={1}+{2}{x}+{3}{x}^{{2}}\)
The auxillary equation is
\(\displaystyle{m}^{{2}}+{4}{m}={0}\)
Step 2
\(\displaystyle{m}^{{2}}+{4}{m}={0}\)
\(\displaystyle{m}{\left({m}+{4}\right)}={0}\)
\(\displaystyle{m}={0}.{m}=-{4}\)
\(\displaystyle{y}_{{c}}={c}_{{1}}{e}^{{{0}{x}}}+{c}_{{2}}{e}^{{-{4}{x}}}\)
\(\displaystyle\Rightarrow{y}_{{c}}={c}_{{1}}+{c}_{{2}}{e}^{{-{4}{x}}}\)
Then \(\displaystyle{y}_{{p}}={A}{x}^{{2}}+{B}{x}\) by using method of undetermined coefficient
\(\displaystyle{y}_{{p}}'={2}{A}{x}+{B}\)
\(\displaystyle{y}_{{p}}\text{}{2}{A}\)
Then \(\displaystyle{y}_{{p}}\text{}{4}{y}_{{p}}'={1}+{2}{x}+{3}{x}^{{2}}\)
\(\displaystyle{2}{A}+{2}{A}{x}+{B}={1}+{2}{x}+{B}{x}^{{2}}\)
\(\displaystyle{\left({2}{A}+{B}\right)}+{2}{A}{x}={3}{x}^{{2}}+{2}{x}+{1}\)
Comparing the coefficients
Step 3

\(\displaystyle{2}{A}={2}.\Rightarrow{A}={1}\)
\(\displaystyle{2}{A}+{B}={1}\Rightarrow{2}{\left({1}\right)}+{B}={1}\)
\(\displaystyle{B}={1}-{2}\)
\(\displaystyle{B}=-{1}\)
Then \(\displaystyle{y}_{{p}}={x}^{{2}}-{x}\)
The general general solution is
\(\displaystyle{y}={y}_{{c}}+{y}_{{p}}\)
\(\displaystyle{y}+{c}_{{1}}+{c}_{{2}}{e}^{{-{4}{x}}}+{x}^{{2}}-{x}\) Isregmined solution

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