Judith McQueen
2022-01-16
Answered

How to solve this differential equation? $y{}^{\u2033}-\frac{{y}^{\prime}}{x}=4{x}^{2}y$

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Debbie Moore

Answered 2022-01-17
Author has **43** answers

Note that we have $xy\prime \prime -y\prime =4{x}^{3}y$ . The LHS in some sense is dimensional consistent and looks something similar to a quotient rule provided we divide by $x}^{2$ . So dividing by $x}^{2$ and doing algebraic manipulations we get $\left(\frac{1}{x}y\prime \right)\prime =4xy$ . Now this looks familiar to some extent.

Rewriting, we get${\left(\frac{{y}^{\prime}}{2x}\right)}^{\prime}=2xy$

Now let$\frac{{y}^{\prime}}{2x}=z\left(x\right)$ . Plug this in and simplify to get ${z}^{\prime}=\frac{{y}^{\prime}}{z}y$

(Replace 2x by$\frac{{y}^{\prime}}{z}$ )

So we have${z}^{2}={y}^{2}+c$

Thus we have now converted a second order differential equation in terms of first order differential equation, viz,

$\frac{1}{2x}\frac{dy}{dx}=\pm \sqrt{{y}^{2}+c}$ where c is a constant.

(You could plug this in and check that this satisfies the second order differential equation.)

We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate c and other constant which will arise after solving the first order differential equation to get$y\left(x\right)=\mathrm{exp}\left({x}^{2}\right)$

(Note that taking$c=0$ we get a simple ode and the solution to which is $y\left(x\right)=y\left(0\right)\mathrm{exp}(\pm {x}^{2}))$ .

EDIT:

The first order ODE can be solved as follows:

CASE 1:

Let$c>0$ , then $c={a}^{2}$

Rearranging, we get

$\frac{dy}{\sqrt{{y}^{2}+{a}^{2}}}=\pm d\left({x}^{2}\right)$

$y=a\mathrm{tan}\left(\theta \right)$ , we get $dy=a{\mathrm{sec}}^{2}\left(\theta \right)d\theta$

Hence, the ode now becomes,

$\mathrm{sec}\left(\theta \right)d\theta =\pm d\left({x}^{2}\right)$

$d\left(\mathrm{log}(\mathrm{sec}\left(\theta \right)+\mathrm{tan}\left(\theta \right))\right)=\pm d\left({x}^{2}\right)$

$\mathrm{log}(\mathrm{sec}\left(\theta \right)+\mathrm{tan}\left(\theta \right)))=\pm ({x}^{2}+k)$

$\mathrm{sec}\left(\theta \right)+\mathrm{tan}\left(\theta \right)=\mathrm{exp}(\pm ({x}^{2}+k))$

Substitute for$\theta$ in terms of y to get,

$\frac{y}{a}\pm \sqrt{1+{\left(\frac{y}{a}\right)}^{2}}=K\mathrm{exp}(\pm {x}^{2})$

CASE 2:

Let$c>0$ , then let $c=-{a}^{2}$

Rearranging, we get

Rewriting, we get

Now let

(Replace 2x by

So we have

Thus we have now converted a second order differential equation in terms of first order differential equation, viz,

(You could plug this in and check that this satisfies the second order differential equation.)

We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate c and other constant which will arise after solving the first order differential equation to get

(Note that taking

EDIT:

The first order ODE can be solved as follows:

CASE 1:

Let

Rearranging, we get

Hence, the ode now becomes,

Substitute for

CASE 2:

Let

Rearranging, we get

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Cassandra Ramirez

Answered 2022-01-18
Author has **30** answers

So

and

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alenahelenash

Answered 2022-01-24
Author has **368** answers

From there, we get the hints of left

Then

The ODE has the simplest form when we choose

The ODE becomes

The auxiliary equation is

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