agreseza

2021-12-31

Find the solution of the following Differential Equations

$\mathrm{sin}\left(\beta y\right)dx=-\beta \mathrm{cos}\left(\beta y\right)dy$

Jenny Sheppard

Beginner2022-01-01Added 35 answers

Solution is

Esther Phillips

Beginner2022-01-02Added 34 answers

Let us discuss what is exact differential equation.

An ordinary differential equation of the form$M(x,y)dx+N(x,y)dy=0$ is exact if

$M(x,y)=\frac{\partial I}{\partial x}$

$N(x,y)=\frac{\partial I}{\partial y}$ NSk$\frac{{\partial}^{2}I}{\partial y\partial x}=\frac{{\partial}^{2}I}{\partial y\partial x}$

$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$

If the differential equation is not exact we can find a integrating factor such that the differential equation becomes exact.

If$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$ is function of x alone i.;e $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=f\left(x\right)$ .

Then the integrating factor is$\mu =\mathrm{exp}(\int f\left(x\right)dx)$

So, the differential equation$\mu M(x,y)dx+\mu N(x,y)dy=0$ becomes exact.

If$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}$ is function of y alone that is $\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=f\left(y\right)$

Then the integrating factor is$\mu =\mathrm{exp}(\int f\left(y\right)dy)$

So, the differential equation$\mu M(x,y)dx+\mu N(x,y)dy=0$ becomes exact.

Given differential equation is$\mathrm{sin}\left(\beta y\right)dx=-\beta \mathrm{cos}\left(\beta y\right)dy$ .

Which is nothing but$\mathrm{sin}\left(\beta y\right)dx+\beta \mathrm{cos}\left(\beta y\right)dy=0$

Let us check if the differential equation is exact or not.

$\frac{\partial \mathrm{sin}\left(\beta y\right)}{\partial y}=\beta \mathrm{cos}\left(\beta y\right)$ . Hence the differential equation is not exact.

$\frac{\partial \beta \mathrm{cos}\left(\beta y\right)}{\partial x}=0$

Let us find the integrating factor.

An ordinary differential equation of the form

If the differential equation is not exact we can find a integrating factor such that the differential equation becomes exact.

If

Then the integrating factor is

So, the differential equation

If

Then the integrating factor is

So, the differential equation

Given differential equation is

Which is nothing but

Let us check if the differential equation is exact or not.

Let us find the integrating factor.