Solve the differential equation. x\frac{dy}{dx}-4y=2x^{4}e^{x}

Jason Farmer 2021-09-16 Answered
Solve the differential equation.
\(\displaystyle{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{4}{y}={2}{x}^{{{4}}}{e}^{{{x}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

berggansS
Answered 2021-09-17 Author has 14506 answers
Step 1
According to the given information, it is required to solve the differential equation.
\(\displaystyle{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{4}{y}={2}{x}^{{{4}}}{e}^{{{x}}}\)
Step 2
First divide the whole differential equation by x to get the linear differential form.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{\frac{{{4}{y}}}{{{x}}}}={2}{x}^{{{3}}}{e}^{{{x}}}\)
Step 3
Now, the general linear differential equation and its solution is:
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{\left({x}\right)}{y}={Q}{\left({x}\right)}\)
solution of differential equation is:
\(\displaystyle{y}{\left({x}\right)}\times{I}{F}=\int{Q}{\left({x}\right)}.{I}{F}{\left.{d}{x}\right.}+{c}\)
where \(\displaystyle{I}{F}={e}^{{\int{P}{\left({x}\right)}{\left.{d}{x}\right.}}}\)
Step 4
Now, solve the given using the above definition.
In the given differential equation is:
\(\displaystyle{P}{\left({x}\right)}={\frac{{-{4}}}{{{x}}}},{Q}{\left({x}\right)}={2}{x}^{{{3}}}{e}^{{{x}}}\)
\(\displaystyle{I}{F}={e}^{{{P}{\left.{d}{x}\right.}}}={e}^{{\int-{\frac{{{4}}}{{{x}}}}{\left.{d}{x}\right.}}}={e}^{{-{4}{\log{{\left({x}\right)}}}}}={e}^{{{\log{{\left({x}^{{-{4}}}\right)}}}}}={x}^{{-{4}}}\)
\(\displaystyle{I}{F}={\frac{{{1}}}{{{x}^{{{4}}}}}}\)
the solution of the given differential equation is:
\(\displaystyle{y}{\left({\frac{{{1}}}{{{x}^{{{4}}}}}}\right)}=\int{2}{x}^{{{3}}}{e}^{{{x}}}{\left({\frac{{{1}}}{{{x}^{{{4}}}}}}\right)}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{4}}}}}}={2}\int{\frac{{{e}^{{{x}}}}}{{{x}}}}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{4}}}}}}={2}{E}_{{{i}}}{\left({x}\right)}+{c}{\left(\int{\frac{{{e}^{{{x}}}}}{{{x}}}}{\left.{d}{x}\right.}={E}_{{{i}}}{\left({x}\right)}+{c}\right)}\)
\(\displaystyle{y}{\left({x}\right)}={2}{x}^{{{4}}}{E}_{{{i}}}{\left({x}\right)}+{c}{x}^{{{4}}}\)
Have a similar question?
Ask An Expert
47
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-02-27
Solve the following differential equations:
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}\)
asked 2021-09-18
Solve the differential equation.
\(\displaystyle{5}\sqrt{{{x}{y}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={1},{x},{y}{>}{0}\)
asked 2021-09-26
Solve differential equation.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\left({x}+{y}-{1}\right)}+{\frac{{{x}+{y}}}{{{\log{{\left({x}+{y}\right)}}}}}}\)
asked 2021-03-21
Find the general solution for each differential equation. Verify that each solution satisfies the original differential equation.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={4}{e}^{{-{3}{x}}}\)
asked 2021-11-15
Solve the diļ¬€erential equation
\(\displaystyle{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}}-{2}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{4}{y}={e}^{{x}}{{\sin}^{{2}}{\left({\frac{{{x}}}{{{2}}}}\right)}}\)
asked 2021-09-28
Solve the equation.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={7}{y}\), and y=1 when z=0.
asked 2021-09-15
Solve the given differential equation by separation of variables.
\(\displaystyle{\left({e}^{{y}}+{1}\right)}^{{2}}{e}^{{-{y}}}{\left.{d}{x}\right.}+{\left({e}^{{x}}+{1}\right)}^{{6}}{e}^{{-{x}}}{\left.{d}{y}\right.}={0}\)
...