Hint: make for the particular solution the ansatz

\(\displaystyle{y}_{{p}}={A}{e}^{{{2}{t}}}{t}^{{2}}\)

\(\displaystyle{y}_{{p}}={A}{e}^{{{2}{t}}}{t}^{{2}}\)

porschomcl

Answered 2021-12-10
Author has **1051** answers

I have no idea what you mean by "sum of A". There is no "A" in the problem. Do you mean that you are looking for a "particular solution" of the form \(\displaystyle{A}{e}^{{{2}{t}}}\)? If so you should NOT be getting A= 0- you should not be getting anything at all for A! Setting \(\displaystyle{y}={A}{e}^{{{2}{t}}},{y}'={2}{A}{e}^{{{2}{t}}}\) and \(\displaystyle{y}{''}={4}{A}{e}^{{{2}{t}}}\) so that \(\displaystyle{y}{''}-{4}{y}+{4}={4}{a}{e}^{{{2}{t}}}-{8}{A}{e}^{{{2}{t}}}+{4}{A}{e}^{{{2}{t}}}={0}\) no matter what A is

The "double root" that Bruce refers to is the double root of the characteristic equation: the associated homogeneous differential equation, \(\displaystyle{y}{''}-{4}{y}'+{4}{y}={0}\), has characteristic equation \(\displaystyle{r}^{{2}}-{4}{y}+{4}={\left({r}-{2}\right)}^{{2}}={0}\) which has the double root r= 2. The general solution to the associated homogeneous equation is \(\displaystyle{y}{\left({t}\right)}={C}_{{1}}{e}^{{{2}{t}}}+{C}_{{2}}{t}{e}^{{{2}{t}}}\). You need to look for a specific solution of the form \(\displaystyle{y}{\left({t}\right)}={A}{t}^{{2}}{e}^{{{2}{t}}}\)

The "double root" that Bruce refers to is the double root of the characteristic equation: the associated homogeneous differential equation, \(\displaystyle{y}{''}-{4}{y}'+{4}{y}={0}\), has characteristic equation \(\displaystyle{r}^{{2}}-{4}{y}+{4}={\left({r}-{2}\right)}^{{2}}={0}\) which has the double root r= 2. The general solution to the associated homogeneous equation is \(\displaystyle{y}{\left({t}\right)}={C}_{{1}}{e}^{{{2}{t}}}+{C}_{{2}}{t}{e}^{{{2}{t}}}\). You need to look for a specific solution of the form \(\displaystyle{y}{\left({t}\right)}={A}{t}^{{2}}{e}^{{{2}{t}}}\)

asked 2021-11-20

Is there any known method to solve such second order non-linear differential equation?

\(\displaystyle{y}{''}_{{n}}-{n}{x}{\frac{{{1}}}{{\sqrt{{{y}_{{n}}}}}}}={0}\)

\(\displaystyle{y}{''}_{{n}}-{n}{x}{\frac{{{1}}}{{\sqrt{{{y}_{{n}}}}}}}={0}\)

asked 2021-12-17

I need to solve the following differential equation

\(\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}\)

with \(\displaystyle{a},{b}{>}{0},{x}\geq{0}\) and \(\displaystyle{y}{\left({0}\right)}={0}\). The power series method will fail since there is a singularity at x=0, while the form of the equation does not conform with the Frobenius method.

\(\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}\)

with \(\displaystyle{a},{b}{>}{0},{x}\geq{0}\) and \(\displaystyle{y}{\left({0}\right)}={0}\). The power series method will fail since there is a singularity at x=0, while the form of the equation does not conform with the Frobenius method.

asked 2021-12-16

Consider the function

\(\displaystyle{f{{\left(\mu\right)}}}=\sum_{{{i}={1}}}{\left({x}_{{{i}}}-\mu\right)}^{{{2}}},\) where

\(\displaystyle{x}_{{{i}}}={i},\ {i}={1},\ {2},\ \cdots,\ {n}\)

What is the first and second derivative os \(\displaystyle{f{{\left(\mu\right)}}}\)?

\(\displaystyle{f{{\left(\mu\right)}}}=\sum_{{{i}={1}}}{\left({x}_{{{i}}}-\mu\right)}^{{{2}}},\) where

\(\displaystyle{x}_{{{i}}}={i},\ {i}={1},\ {2},\ \cdots,\ {n}\)

What is the first and second derivative os \(\displaystyle{f{{\left(\mu\right)}}}\)?

asked 2021-12-16

I have this second-order ode equation:

\(\displaystyle{y}{''}-{4}{y}'+{13}{y}={0}\)

I've identified it as x missing case as \(\displaystyle{y}{''}={f{{\left({y}',{y}\right)}}}={4}{y}'-{13}{y}\), so I'm substituting witth:

\(\displaystyle{y}'={P},\ {y}{''}={P}{\frac{{{\left.{d}{y}\right.}^{{2}}}}{{{d}^{{2}}{x}}}}={f{{\left({P},{y}\right)}}}={4}{P}-{13}{y}\)

At this point I have \(\displaystyle{P}{\frac{{{d}{p}}}{{{\left.{d}{y}\right.}}}}={4}{P}-{13}{y}\), which seems a non-linear first-order ODE. This is currently beyond the scope of my course, so I'm unsure if I should continue and search online for solving techniques, or did I already do something wrong?

\(\displaystyle{y}{''}-{4}{y}'+{13}{y}={0}\)

I've identified it as x missing case as \(\displaystyle{y}{''}={f{{\left({y}',{y}\right)}}}={4}{y}'-{13}{y}\), so I'm substituting witth:

\(\displaystyle{y}'={P},\ {y}{''}={P}{\frac{{{\left.{d}{y}\right.}^{{2}}}}{{{d}^{{2}}{x}}}}={f{{\left({P},{y}\right)}}}={4}{P}-{13}{y}\)

At this point I have \(\displaystyle{P}{\frac{{{d}{p}}}{{{\left.{d}{y}\right.}}}}={4}{P}-{13}{y}\), which seems a non-linear first-order ODE. This is currently beyond the scope of my course, so I'm unsure if I should continue and search online for solving techniques, or did I already do something wrong?

asked 2021-12-10

I am trying to solve the following:

\(\displaystyle{y}{''}+{4}{y}'={\tan{{\left({t}\right)}}}\)

I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:

\(\displaystyle{u}_{{1}}=\int{\frac{{{\tan{{t}}}{\cos{{2}}}{t}}}{{{2}}}}\)

I am stuck here

\(\displaystyle{y}{''}+{4}{y}'={\tan{{\left({t}\right)}}}\)

I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:

\(\displaystyle{u}_{{1}}=\int{\frac{{{\tan{{t}}}{\cos{{2}}}{t}}}{{{2}}}}\)

I am stuck here

asked 2021-11-19

We have the following differential equation

\(\displaystyle\cup{''}={1}+{\left({u}'\right)}^{{2}}\)

i found that the general solution of this equation is

\(\displaystyle{u}={d}{\text{cosh}{{\left({\left({x}-\frac{{b}}{{d}}\right)}\right.}}}\)

where b and d are constats

Please how we found this general solution?

\(\displaystyle\cup{''}={1}+{\left({u}'\right)}^{{2}}\)

i found that the general solution of this equation is

\(\displaystyle{u}={d}{\text{cosh}{{\left({\left({x}-\frac{{b}}{{d}}\right)}\right.}}}\)

where b and d are constats

Please how we found this general solution?

asked 2021-11-19

I need some hints for solving \(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)

I noticed that the left hand side is close to \(\displaystyle{\left({y}{y}'\right)}'\):

\(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{y}{y}{''}+{\left({y}'\right)}^{{2}}-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{\left({y}{y}'\right)}'-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)

But I don't know how to continue expressing the terms as derivatives of some functions.

I noticed that the left hand side is close to \(\displaystyle{\left({y}{y}'\right)}'\):

\(\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{y}{y}{''}+{\left({y}'\right)}^{{2}}-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{\left({y}{y}'\right)}'-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\)

But I don't know how to continue expressing the terms as derivatives of some functions.