Question

Find the particular solution of the differential equation
${{e}^{x}}\sqrt{1-{{y}^{2}}}dx+\dfrac{y}{x}dy=0$ , given that $y=1$ when $x=0$.

Answer 1

The given homogeneous equation is separable under $x$ and $y$ , we will separate the terms, the terms involving $x$ in the Left Hand Side and terms involving $y$ on the Right Hand Side and then apply integration on both the sides of the equation, and add integration constant . To find the value of integration constant put $x=0,y=1$ and get the desired solution.

Complete step by step answer:
We are given a first order homogeneous separable differential equation, so first we will separate the differential equation as follows,
$\begin{aligned} & {{e}^{x}}\sqrt{1-{{y}^{2}}}dx=-\dfrac{y}{x}dy \\\ & \Rightarrow x{{e}^{x}}dx=-\dfrac{y}{\sqrt{1-{{y}^{2}}}}dy \\\ \end{aligned}$
Now, that we have separated the terms,
Integrate on both the sides and get,
$-\int{x{{e}^{x}}dx=\int{\dfrac{y}{\sqrt{1-{{y}^{2}}}}dy}}+C$
; where $C$ is integration constant.
Now, we know the formula of By-parts in integration as
$\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{ \dfrac{d}{dx}f\left( x \right)\int{\left( g\left( x \right)dx \right)dx} }$
; where we have to choose $f\left( x \right)$ and $g\left( x \right)$ wisely, so that the integration does not become complicated.
$\begin{aligned} & -[x.{{e}^{x}}-\int_{{}}^{{}}{{{e}^{x}}dx]}=\int{\dfrac{y}{\sqrt{1-{{y}^{2}}}}dy}+C \\\ & -x{{e}^{x}}+{{e}^{x}}=\int{\dfrac{y}{\sqrt{1-{{y}^{2}}}}dy}+C \\\ \end{aligned}$
Applying this rule to integrate the integration involving $x$ , we get
$x{{e}^{x}}-{{e}^{x}}=-\int{\dfrac{y}{\sqrt{1-{{y}^{2}}}}dy}+C$
Now,
Let, $1-{{y}^{2}}=t$
Differentiating both sides, we get
$\begin{aligned} & -2ydy=dt \\\ & \Rightarrow -ydy=\dfrac{dt}{2} \\\ \end{aligned}$
So, converting the integral in terms of $t$ , we get,
$\begin{aligned} & x{{e}^{x}}-{{e}^{x}}=\dfrac{1}{2}\int{\dfrac{1}{\sqrt{t}}dt}+C \\\ & \Rightarrow x{{e}^{x}}-{{e}^{x}}=\sqrt{t}+C \\\ \end{aligned}$
Now, putting the value of $t$ in the last step, we get
$x{{e}^{x}}-{{e}^{x}}=\sqrt{1-{{y}^{2}}}+C$
Hence, the general solution of the given differential equation is
$x{{e}^{x}}-{{e}^{x}}=\sqrt{1-{{y}^{2}}}+C$
Now, since we have to find the particular solution, we have to find the value of $C$ , such that it satisfies the solution passing through $x=0,y=1$
Now, putting the value of $x$ and $y$ in the solution, we get,
$\begin{aligned} & 0{{e}^{0}}-{{e}^{0}}=\sqrt{1-{{1}^{2}}}+C \\\ & \Rightarrow -1=C \\\ \end{aligned}$
Now, putting the value of $C$ in the solution to get the particular solution,
$x{{e}^{x}}-{{e}^{x}}=\sqrt{1-{{y}^{2}}}-1$

Hence ,
The particular solution the given differential equation satisfying $x=0,y=1$ is
$x{{e}^{x}}-{{e}^{x}}=\sqrt{1-{{y}^{2}}}-1$

Note: The possibility of mistake here is when we use by parts in a step if we choose first function as ${{e}^{x}}$ , Choosing this as the first function would be a complete disaster, you would be using by parts again and again, and still not getting an answer. So, whenever you use parts, always choose the first and second function in a way that in a few steps you could have integral value.
Since this problem is very complicated we can apply the integration by parts for this we have to apply the ILATE rule means while integrating by parts we have to give the preference which one we will integrate first so this rule is used in integration by parts, we have learned when the product of two functions are given to us then we apply the required formula. The integral of the two functions is taken, by considering the left term as first function and second term as the second function. This method is called the ILATE rule. Where I mean inverse, L means Logarithmic, A means Arithmetic, T means Trigonometric, E means Exponential.