guringpw

2021-12-28

Evaluate the following differential equations:

Integrating Factor by Formula

$(2{y}^{2}+2y+4{x}^{2})dx+(2xy+x)dy=0$

Integrating Factor by Formula

deginasiba

Beginner2021-12-29Added 31 answers

Finally if for given D.E is

Buck Henry

Beginner2021-12-30Added 33 answers

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$\frac{1}{N}(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})=\frac{1}{2xy+x}(4y+2-2y-1)$

$=\frac{1}{x(2y+1)}(2y+1)$

$=\frac{1}{x}$ , 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

$x(2{y}^{2}+2y+4{x}^{2})dx+x(2xy+x)dy=0$

$(2x{y}^{2}+2xy+4{x}^{3})dx+(2{x}^{2}y+{x}^{2})dy=0$

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

${\int}_{\text{Treating y as constant}}{M}_{1}dx+\int \left(\text{terms in}{N}_{1}\text{not containing x}\right)dy=c$

${\int}_{\text{Treating y as constant}}(2x{y}^{2}+2xy+4{x}^{3})dx=c$

$2{y}^{2}\cdot \frac{{x}^{2}}{2}+2y\cdot \frac{{x}^{2}}{2}+4\cdot \frac{{x}^{4}}{4}=c$

Comparing (i) with

Here

Here

So,

We have

Multiplying (i) by x , we get

Here

which must be an exact equation and so its solution as usual is

karton

Expert2022-01-09Added 613 answers

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