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

$\mathrm{sin}\left(\beta y\right)dx=-\beta \mathrm{cos}\left(\beta y\right)dy$
$dx=-\beta \frac{\mathrm{cos}\left(\beta y\right)}{\mathrm{sin}\left(\beta y\right)}dy$
$\int dx=-\beta \int \mathrm{cot}\left(\beta y\right)dy$
$x=-\beta \frac{\mathrm{ln}\left(\beta y\right)}{\beta }+c$
$x=-\frac{\mathrm{ln}\left(\beta y\right)}{1}+c$
Solution is
$x+\mathrm{ln}\left(\beta y\right)+c=0$
$C\to$\ integration constant.

Esther Phillips

Let us discuss what is exact differential equation.
An ordinary differential equation of the form $M\left(x,y\right)dx+N\left(x,y\right)dy=0$ is exact if
$M\left(x,y\right)=\frac{\partial I}{\partial x}$
$N\left(x,y\right)=\frac{\partial I}{\partial y}$NSk$\frac{{\partial }^{2}I}{\partial y\partial x}=\frac{{\partial }^{2}I}{\partial y\partial x}$
$\frac{\partial M\left(x,y\right)}{\partial y}=\frac{\partial N\left(x,y\right)}{\partial x}$
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 $\mu =\mathrm{exp}\left(\int f\left(x\right)dx\right)$
So, the differential equation $\mu M\left(x,y\right)dx+\mu N\left(x,y\right)dy=0$ 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 $\mu =\mathrm{exp}\left(\int f\left(y\right)dy\right)$
So, the differential equation $\mu M\left(x,y\right)dx+\mu N\left(x,y\right)dy=0$ 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.