Question

Solve the question $xdy-ydx=\sqrt{{{x}^{2}}-{{y}^{2}}}dy$ is?

Answer 1

Hint: Observe the given equation to get the type of it. For homogeneous differential equation, the following relation I followed by the equation, given as
$F\left( \lambda x,\lambda y \right)={{\lambda }^{n}}F\left( x,y \right)$ , where $F\left( x,y \right)$ is the given differential equation. Replace x by vy, using another variable ‘V’. Use the following formula $\int{\dfrac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}}={{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right)}$.

Complete step-by-step answer:
Given differential equation in the question is
$xdy-ydx=\sqrt{{{x}^{2}}-{{y}^{2}}}dy$ ……………………………….(i)
Now, we can write the above differential equation as
$xdy-\sqrt{{{x}^{2}}-{{y}^{2}}}dy=ydx$
On dividing the whole equation by ‘dy’, we get
$x-\sqrt{{{x}^{2}}-{{y}^{2}}}=y\dfrac{dx}{dy}$
$\Rightarrow \dfrac{dx}{dy}=\dfrac{x-\sqrt{{{x}^{2}}-{{y}^{2}}}}{y}$ …......................................(ii)
Now, we need to observe the formed differential equation as mentioned above. As, we get that the equation is not of the form of separate type or linear one as well.
So, let us check the given equation is representing a homogeneous equation or not.
So, any differentiation equal will be of homogeneous type if
$F\left( \lambda x,\lambda y \right)={{\lambda }^{n}}F\left( x,y \right)$ …………………………………(iii)
So, let us suppose the given differentiation equation (ii) as $F\left( x,y \right)$. So, we get
$F\left( x,y \right)=\dfrac{x-\sqrt{{{x}^{2}}-{{y}^{2}}}}{y}$
Hence, we get
$\Rightarrow F\left( \lambda x,\lambda y \right)=\dfrac{\lambda x-\sqrt{{{\left( \lambda x \right)}^{2}}-{{\left( \lambda y \right)}^{2}}}}{\lambda y}$
$\Rightarrow \dfrac{\lambda x-\sqrt{{{\lambda }^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)}}{\lambda y}$
$\Rightarrow \dfrac{\lambda \left[ x-\sqrt{\left( {{x}^{2}}-{{y}^{2}} \right)} \right]}{\lambda y}$
$\Rightarrow F\left( \lambda x,\lambda y \right)={{\lambda }^{0}}F\left( x,y \right)$
Hence, we get that given differential equation is representing a homogeneous equation.
So, let $x=Vy$ ……………………………..(iv)
Differentiate both sides of the above expression and apply the multiplication rule of differentiation, which is given as
$\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ ……………………(v)
So, we get the equation on differentiating equation (v) w.r.t ‘y’ as
$\dfrac{dx}{dy}=v\dfrac{dy}{dy}+y\dfrac{dv}{dy}$
$\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$ …………………………………(vi)
Hence, we can get equation (ii), with the help of equation (iv) and (vi) as
$\Rightarrow v+y\dfrac{dv}{dy}=\dfrac{vy-\sqrt{{{\left( vy \right)}^{2}}-{{y}^{2}}}}{y}$
$\Rightarrow v+y\dfrac{dv}{dy}=\dfrac{y\left( v-\sqrt{{{v}^{2}}-1} \right)}{y}$
$\Rightarrow v+y\dfrac{dv}{dy}=v-\sqrt{{{v}^{2}}-1}$
$\Rightarrow y\dfrac{dv}{dy}=-\sqrt{{{v}^{2}}-1}$
Now, we can separate the variables v and y as
$\int{\dfrac{dv}{\sqrt{{{v}^{2}}-1}}=\int{\dfrac{-dy}{y}}}$
Now, we know
$\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}={{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right)}$
$\Rightarrow \int{\dfrac{dx}{x}={{\log }_{e}}x}$
Hence, we get
$\Rightarrow \int{\dfrac{dv}{\sqrt{{{v}^{2}}-1}}=-\int{\dfrac{1}{y}dy}}$
$\Rightarrow {{\log }_{e}}\left( v+\sqrt{{{v}^{2}}-1} \right)=-\left( {{\log }_{e}}y \right)+c$
$\Rightarrow {{\log }_{e}}\left( v+\sqrt{{{v}^{2}}-1} \right)=-{{\log }_{e}}y+c$
Put $v=\dfrac{x}{y}$ from the equation (iv). So, we get
$\Rightarrow {{\log }_{e}}\left( \dfrac{x}{y}+\sqrt{{{\left( \dfrac{x}{y} \right)}^{2}}-1} \right)=-{{\log }_{e}}y+c$
$\Rightarrow {{\log }_{e}}\left( \dfrac{x}{y}+\sqrt{{{\left( \dfrac{x}{y} \right)}^{2}}-1} \right)+{{\log }_{e}}y=c$ ……………………………(vii)
We can replace c by logarithm constant ${{\log }_{e}}c$ as well, as ${{\log }_{e}}c$ will also act as a constant.
And we know, ${{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}mn$
So, we get equation (vii) as
$\Rightarrow {{\log }_{e}}\left( \left( \dfrac{x}{y}+\sqrt{{{\left( \dfrac{x}{y} \right)}^{2}}-1} \right)\times y \right)={{\log }_{e}}c$
$\Rightarrow {{\log }_{e}}\left( \left( \dfrac{x}{y}+\sqrt{\dfrac{{{x}^{2}}-{{y}^{2}}}{{{y}^{2}}}} \right)\times y \right)={{\log }_{e}}c$
\[\Rightarrow {{\log }_{e}}\left( \dfrac{x+\sqrt{{{x}^{2}}-{{y}^{2}}}}{y}\times y \right)={{\log }_{e}}c\]
\[\Rightarrow {{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-{{y}^{2}}} \right)={{\log }_{e}}c\]
Now, we can compare the above equation on both sides. And hence, we get
\[x+\sqrt{{{x}^{2}}-{{y}^{2}}}=c\]
Hence, the solution of the differential equation is \[x+\sqrt{{{x}^{2}}-{{y}^{2}}}=c\].

Note: One may use \[y=vx\] substitution as well, and apply the same rule of differentiation to get the value of \[\dfrac{dx}{dy}\]. So, it can be another approach.
Be clear with the formula \[\int{\dfrac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right)\] to solve this kind of problems. Otherwise we will not be able to solve this kind of question fully.
Don’t confuse the taken constant term i.e. \[{{\log }_{e}}c\]. It is done in the solution to get the solution of the differential equation in simplified form. So, don’t confuse yourself with that step.