guringpw

2021-12-28

Evaluate the following differential equations:
Integrating Factor by Formula
$\left(2{y}^{2}+2y+4{x}^{2}\right)dx+\left(2xy+x\right)dy=0$

### Answer & Explanation

deginasiba

$Mdx+Ndy=0$

$\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$ equation is not exact.
$IF=\frac{{M}_{y}-{N}_{x}}{N}=\frac{\left(4y+2\right)-\left(2y+1\right)}{2xy+x}$
$=\frac{4y-2y+2-1}{x\left(2y+1\right)}=\frac{2y+1}{x\left(2y+1\right)}=\frac{1}{x}$
$IF=\frac{{M}_{y}-{N}_{x}}{N}=\frac{1}{x}=f\left(x\right)$
Finally if for given D.E is $=\frac{1}{x}$

Buck Henry

Explanation:
Comparing (i) with $Mdx+Ndy=0$
Here $M=2{y}^{2}+2y+4{x}^{2}\text{and}N=2xy+x$
Here $\frac{\partial M}{\partial y}=4y+2\text{and}\frac{\partial N}{\partial x}=2y+1$
So, $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
We have $\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=\frac{1}{2xy+x}\left(4y+2-2y-1\right)$
$=\frac{1}{x\left(2y+1\right)}\left(2y+1\right)$
$=\frac{1}{x}$, which is a function of x alone
$\therefore I.Fof\left(i\right)={e}^{\int \frac{1}{x}dx}={e}^{\mathrm{log}x}=x$
Multiplying (i) by x , we get
$x\left(2{y}^{2}+2y+4{x}^{2}\right)dx+x\left(2xy+x\right)dy=0$
$\left(2x{y}^{2}+2xy+4{x}^{3}\right)dx+\left(2{x}^{2}y+{x}^{2}\right)dy=0$
Here ${M}_{1}=2x{y}^{2}+2xy+4{x}^{3}\text{and}{N}_{1}=2{x}^{2}y+{x}^{2}$
which must be an exact equation and so its solution as usual is
${\int }_{\text{Treating y as constant}}{M}_{1}dx+\int \left(\text{terms in}{N}_{1}\text{not containing x}\right)dy=c$
${\int }_{\text{Treating y as constant}}\left(2x{y}^{2}+2xy+4{x}^{3}\right)dx=c$
$2{y}^{2}\cdot \frac{{x}^{2}}{2}+2y\cdot \frac{{x}^{2}}{2}+4\cdot \frac{{x}^{4}}{4}=c$

karton