# Find the solution of the following Differential Equations \sin(\beta y)dx =-\beta

Find the solution of the following Differential Equations
$$\displaystyle{\sin{{\left(\beta{y}\right)}}}{\left.{d}{x}\right.}=-\beta{\cos{{\left(\beta{y}\right)}}}{\left.{d}{y}\right.}$$

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Jenny Sheppard
$$\displaystyle{\sin{{\left(\beta{y}\right)}}}{\left.{d}{x}\right.}=-\beta{\cos{{\left(\beta{y}\right)}}}{\left.{d}{y}\right.}$$
$$\displaystyle{\left.{d}{x}\right.}=-\beta{\frac{{{\cos{{\left(\beta{y}\right)}}}}}{{{\sin{{\left(\beta{y}\right)}}}}}}{\left.{d}{y}\right.}$$
$$\displaystyle\int{\left.{d}{x}\right.}=-\beta\int{\cot{{\left(\beta{y}\right)}}}{\left.{d}{y}\right.}$$
$$\displaystyle{x}=-\beta{\frac{{{\ln{{\left(\beta{y}\right)}}}}}{{\beta}}}+{c}$$
$$\displaystyle{x}=-{\frac{{{\ln{{\left(\beta{y}\right)}}}}}{{{1}}}}+{c}$$
Solution is
$$\displaystyle{x}+{\ln{{\left(\beta{y}\right)}}}+{c}={0}$$
$$\displaystyle{C}\rightarrow$$\ integration constant.
###### Not exactly what you’re looking for?
Esther Phillips
Let us discuss what is exact differential equation.
An ordinary differential equation of the form $$\displaystyle{M}{\left({x},{y}\right)}{\left.{d}{x}\right.}+{N}{\left({x},{y}\right)}{\left.{d}{y}\right.}={0}$$ is exact if
$$\displaystyle{M}{\left({x},{y}\right)}={\frac{{\partial{I}}}{{\partial{x}}}}$$
$$\displaystyle{N}{\left({x},{y}\right)}={\frac{{\partial{I}}}{{\partial{y}}}}$$ NSk $$\displaystyle{\frac{{\partial^{{{2}}}{I}}}{{\partial{y}\partial{x}}}}={\frac{{\partial^{{{2}}}{I}}}{{\partial{y}\partial{x}}}}$$
$$\displaystyle{\frac{{\partial{M}{\left({x},{y}\right)}}}{{\partial{y}}}}={\frac{{\partial{N}{\left({x},{y}\right)}}}{{\partial{x}}}}$$
If the differential equation is not exact we can find a integrating factor such that the differential equation becomes exact.
If $$\displaystyle{\frac{{{\frac{{\partial{M}}}{{\partial{y}}}}-{\frac{{\partial{N}}}{{\partial{x}}}}}}{{{N}}}}$$ is function of x alone i.;e $$\displaystyle{\frac{{{\frac{{\partial{M}}}{{\partial{y}}}}-{\frac{{\partial{N}}}{{\partial{x}}}}}}{{{N}}}}={f{{\left({x}\right)}}}$$.
Then the integrating factor is $$\displaystyle\mu={\exp{{\left(\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\right)}}}$$
So, the differential equation $$\displaystyle\mu{M}{\left({x},{y}\right)}{\left.{d}{x}\right.}+\mu{N}{\left({x},{y}\right)}{\left.{d}{y}\right.}={0}$$ becomes exact.
If $$\displaystyle{\frac{{{\frac{{\partial{N}}}{{\partial{x}}}}-{\frac{{\partial{M}}}{{\partial{y}}}}}}{{{M}}}}$$ is function of y alone that is $$\displaystyle{\frac{{{\frac{{\partial{N}}}{{\partial{x}}}}-{\frac{{\partial{M}}}{{\partial{y}}}}}}{{{M}}}}={f{{\left({y}\right)}}}$$
Then the integrating factor is $$\displaystyle\mu={\exp{{\left(\int{f{{\left({y}\right)}}}{\left.{d}{y}\right.}\right)}}}$$
So, the differential equation $$\displaystyle\mu{M}{\left({x},{y}\right)}{\left.{d}{x}\right.}+\mu{N}{\left({x},{y}\right)}{\left.{d}{y}\right.}={0}$$ becomes exact.
Given differential equation is $$\displaystyle{\sin{{\left(\beta{y}\right)}}}{\left.{d}{x}\right.}=-\beta{\cos{{\left(\beta{y}\right)}}}{\left.{d}{y}\right.}$$.
Which is nothing but $$\displaystyle{\sin{{\left(\beta{y}\right)}}}{\left.{d}{x}\right.}+\beta{\cos{{\left(\beta{y}\right)}}}{\left.{d}{y}\right.}={0}$$
Let us check if the differential equation is exact or not.
$$\displaystyle{\frac{{\partial{\sin{{\left(\beta{y}\right)}}}}}{{\partial{y}}}}=\beta{\cos{{\left(\beta{y}\right)}}}$$. Hence the differential equation is not exact.
$$\displaystyle{\frac{{\partial\beta{\cos{{\left(\beta{y}\right)}}}}}{{\partial{x}}}}={0}$$
Let us find the integrating factor.
$$\displaystyle{\frac{{{\frac{{\partial{M}}}{{\partial{y}}}}-{\frac{{\partial{N}}}{{\partial{x}}}}}}{{{N}}}}={\frac{{\beta{\cos{{\left(\beta{y}\right)}}}-{0}}}{{\beta{\cos{{\left(\beta{y}\right)}}}}}}={1}$$ which is function of x alone .
Therefore, the integrating factor is $$\displaystyle{\exp{{\left(\int{1}{\left.{d}{x}\right.}\right)}}}={\exp{{\left({x}\right)}}}$$
The exact differential equation is $$\displaystyle{e}^{{{x}}}{\sin{{\left(\beta{y}\right)}}}{\left.{d}{x}\right.}+{e}^{{{x}}}\beta{\cos{{\left(\beta{y}\right)}}}{\left.{d}{y}\right.}={0}$$
Let us see how to solve an exact differential equation.
Let I be the integral of differential equation.
$$\displaystyle{I}=\int{M}{\left({x},{y}\right)}$$ treating y as independent variable.
So, $$\displaystyle{I}=\int{M}{\left({x},{y}\right)}{\left.{d}{x}\right.}+{f{{\left({y}\right)}}}$$
$$\displaystyle{\frac{{\partial{I}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left(\int{M}{\left({x},{y}\right)}{\left.{d}{x}\right.}\right)}+{f}'{\left({y}\right)}$$
$$\displaystyle{f}'{\left({y}\right)}={N}{\left({x},{y}\right)}$$
$$\displaystyle{f{{\left({y}\right)}}}=\int{N}{\left({x},{y}\right)}{\left.{d}{y}\right.}+{c}$$
$$\displaystyle{I}=\int{e}^{{{x}}}{\sin{{\left(\beta{y}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle={e}^{{{x}}}{\sin{{\left(\beta{y}\right)}}}+{f{{\left({y}\right)}}}$$
$$\displaystyle{\frac{{\partial{I}}}{{\partial{y}}}}={e}^{{{x}}}\beta{\left({\cos{{\left(\beta{y}\right)}}}\right)}+{f}'{\left({y}\right)}$$
$$\displaystyle{e}^{{{x}}}\beta{\cos{\beta}}{y}+{f}'{\left({y}\right)}={e}^{{{x}}}\beta{\cos{\beta}}{y}$$
$$\displaystyle{f}'{\left({y}\right)}={0}$$
$$\displaystyle{f{{\left({y}\right)}}}={c}$$
Hence, the integral of differential equation is $$\displaystyle{e}^{{{x}}}{\sin{{\left(\beta{y}\right)}}}+{c}={0}$$.
karton

$$\begin{array}{} \frac{\partial^{2}I}{\partial y \partial x}=\frac{\partial^{2}I}{\partial y \partial x} \\\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} \\If\ \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\ is\ function\ of\ x\ alone\ i.e\ \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=f(x). \\Then\ the\ integrating\ factor\ is\ \mu=exp(\int f(x)dx) \\So,\ the\ differential\ equation\ \mu M(x,y)dx+\mu N(x,y)dy=0\ becomes\ exact. \\If\ \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M} is function of y alone that is \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=f(y) Then\ the\ integrating\ factor\ is\\mu=exp(\int f(y)dy) \\So,\ the\ differential\ equation\ \mu M(x,y)dx+\mu N(x,y)dy=0\ becomes\ exact. \\Given\ differential\ equation\ is\ \sin( \beta y)dx=- \beta \cos(\beta y)dy \\Which\ is\ nothing\ but\ \sin (\beta y)dx+\beta \cos (\beta y)dy=0 \\\frac{\partial \sin (\beta y)}{\partial y}=\beta \cos (\beta y).\ Hence\ the\ differential\ equation\ is\ not\ exact. \\\frac{\partial \beta \cos (\beta y)}{\partial x}=0 \\\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{\beta \cos (\beta y)-0}{\beta \cos (\beta y)}=1 which\ is\ function\ of\ x\ alone . \\Therefore,\ the\ integrating\ factor\ is\ exp(\int 1dx)=exp (x) \\The\ exact\ differential\ equation\ is\ e^{x} \sin (\beta y)dx+e^{x} \beta \cos (\beta y)dy=0 \\So\, I=\int M(x,y)dx+f(y) \\\frac{\partial I}{\partial y}=\frac{\partial}{\partial y} (\int M(x,y)dx)+f'(y) \\f'(y)=N(x,y) \\f(y)=\int N(x,y)dy+c \\I=\int e^{x} \sin (\beta y)dx \\=e^{x} \sin (\beta y)+f(y) \\\frac{\partial I}{\partial y}=e^{x} \beta (\cos (\beta y))+f'(y) \\e^{x} \beta \cos \beta y+f'(y)=e^{x} \beta \cos \beta y \\f'(y)=0 \\f(y)=c \\Answer:\ Ke^{x} \sin (\beta y)+c=0. \end{array}$$