Marenonigt

2021-12-26

Differential equations in the following form are called Bernoulli Equations.
Find the solution for the following initial value problems and find the interval of validity for the solution.

MoxboasteBots5h

Step 1
The given Bernoulli Equation is
${y}^{\prime }+\frac{4}{3}y={x}^{3}{y}^{2}$ (1)
Compare with,
${y}^{\prime }+P\left(x\right)y=Q\left(x\right){y}^{n}$
Step 2
$⇒P\left(x\right)=\frac{4}{x},Q\left(x\right)={x}^{3}\text{and}n=2$
Now, the integrating factor $I.F.={e}^{\int \left(1-n\right)P\left(x\right)dx}$
$⇒I.F.={e}^{\int \left(1-2\right)×\frac{4}{x}dx}$
$⇒I.F.={e}^{-4}\int \frac{1}{x}dx$
$⇒I.F.={e}^{-4\mathrm{ln}\left(x\right)}$
$⇒I.F.={e}^{\mathrm{ln}\left(x\right)-4}$
$⇒I.F.={x}^{-4}$
$⇒I.F.=\frac{1}{{x}^{4}}$
$⇒I\left(x\right)=\frac{1}{{x}^{4}}$
Step 3
The solution of the given differential equation is,
${y}^{1-n}=\frac{1}{l\left(x\right)}\left[\int \left(1-n\right)Q\left(x\right)l\left(x\right)dx+c\right]$
$⇒{y}^{1-2}=\frac{1}{{x}^{4}}\left[\int \left(1-2\right){x}^{3}×\frac{1}{{x}^{4}}dx+c\right]$
$⇒{y}^{-1}=\frac{1}{{x}^{4}}\left[-\int \frac{1}{x}dx+c\right]$
$⇒\frac{{x}^{4}}{y}=-\mathrm{ln}|x|+c$
$⇒{x}^{4}+y\mathrm{ln}|x|=cy$ (2)
$\therefore y\left(2\right)=-1$
$⇒\text{put}x=2\text{and}y=-1$
$⇒{\left(2\right)}^{4}+\left(-1\right)\mathrm{ln}|2|=-c$
$⇒c=\mathrm{ln}\left(2\right)-16$
From (2)

Mason Hall

${y}^{\prime }+\frac{4}{x}y-{x}^{3}{y}^{2}=0$
${y}^{\prime }+\frac{4}{x}y={x}^{3}{y}^{2}$
$\frac{1}{{y}^{2}}{y}^{\prime }+\frac{4}{x}\frac{1}{y}={x}^{3}$
Suppose, $\frac{1}{y}=z$
$\frac{1}{{y}^{2}}{y}^{\prime }=-{z}^{\prime }$
So, the DE bevomes
$-{z}^{\prime }+\frac{4}{x}z={x}^{3}$
${z}^{\prime }-\frac{4}{x}z=-{x}^{3}$
So, the integrating factor is
$I.F.={e}^{\int -\frac{4}{3}dx}={e}^{-4\mathrm{ln}x}={e}^{{\mathrm{ln}x}^{-4}}={x}^{-4}$
So, the solution is
$z\left[I.F.\right]=\int Q\left\{I.F.\right\}dx$
$z\left[{x}^{-4}\right]=\int {x}^{3}{x}^{-4}dx$
$z\left[{x}^{-4}\right]=\int {x}^{-1}dx$
$z\left[{x}^{-4}\right]=\mathrm{ln}x+c$
$z=\left(\mathrm{ln}x+c\right){x}^{4}$
$\frac{1}{y}=\left(\mathrm{ln}x+c\right){x}^{4}$
$y=\frac{1}{\left(\mathrm{ln}x+c\right){x}^{4}}$

karton