Question

Consider the differential equation, ${{y}^{2}}dx+\left( x-\dfrac{1}{y} \right)dy=0.$ If the value of $y$ is 1 when $x=1$, then the value of $x$ for which $y=2$, is?

Answer 1

Hint: We start solving this problem by converting the given differential equation into the form $\dfrac{dx}{dy}+x.P\left( y \right)=Q\left( y \right)$ . Then we find the integrating factor of the differential equation by using the formula ${{e}^{\int{P\left( y \right)dy}}}$ . Then we solve the differential equation by using the formula \[x.{{e}^{\int{P\left( y \right)dy}}}=\int{Q\left( y \right).{{e}^{\int{P\left( y \right)dy}}}dy}+C\] . Then we use the formula $\int{f\left( x \right)g\left( x \right)dx=}f\left( x \right)\int{g\left( x \right)dx}-\int{{f}'\left( x \right)\left( \int{g\left( x \right)dx} \right)}dx$ . Then we get the solution of the given integral. We Substitute $x=1$ and $y=1$ to get the value of the constant term in the solution. Then we substitute $y=2$ in the final equation to get the value of $x$.

Complete step by step answer:
Let us consider the given differential equation,
${{y}^{2}}dx+\left( x-\dfrac{1}{y} \right)dy=0$
Now, we divide the above differential equation by $dy$ on both sides. Then we get,
$\begin{align}
  & {{y}^{2}}\dfrac{dx}{dy}+\left( x-\dfrac{1}{y} \right)\dfrac{dy}{dy}=\dfrac{0}{dy} \\
 & \Rightarrow {{y}^{2}}\dfrac{dx}{dy}+\left( x-\dfrac{1}{y} \right)=0 \\
 & \Rightarrow {{y}^{2}}\dfrac{dx}{dy}+x=\dfrac{1}{y} \\
\end{align}$
Now, let us divide the above differential equation by ${{y}^{2}}$ on both the sides, we get,
$\dfrac{dx}{dy}+\dfrac{x}{{{y}^{2}}}=\dfrac{1}{{{y}^{3}}}$
By comparing the above differential equation with $\dfrac{dx}{dy}+x.P\left( y \right)=Q\left( y \right)$ , we get, $P\left( y \right)=\dfrac{1}{{{y}^{2}}}$ and $Q\left( y \right)=\dfrac{1}{{{y}^{3}}}$.
Now, let us consider the formula of integrating factor, ${{e}^{\int{P\left( y \right)dy}}}$
By using the above formula, we find the integrating factor. Then we get,
$\begin{align}
  & {{e}^{\int{P\left( y \right)dy}}} \\
 & ={{e}^{\int{\dfrac{1}{{{y}^{2}}}dy}}} \\
 & ={{e}^{\dfrac{-1}{y}}} \\
\end{align}$
So, the integrating factor is ${{e}^{\dfrac{-1}{y}}}$.
Now, let us consider the formula for the solution of a differential equation of the form $\dfrac{dx}{dy}+x.P\left( y \right)=Q\left( y \right)$ , that is, \[x.{{e}^{\int{P\left( y \right)dy}}}=\int{Q\left( y \right).{{e}^{\int{P\left( y \right)dy}}}dy}+C\]
By using the above formula, we get,
\[\begin{align}
  & x.{{e}^{\int{P\left( y \right)dy}}}=\int{Q\left( y \right).{{e}^{\int{P\left( y \right)dy}}}dy}+C \\
 & x.{{e}^{\dfrac{-1}{y}}}=\int{\dfrac{1}{{{y}^{3}}}.{{e}^{\dfrac{-1}{y}}}dy}+C \\
\end{align}\]
Let us consider $t=\dfrac{-1}{y}$ .
By differentiating on both the sides gives us
$dt=\dfrac{-1}{{{y}^{2}}}dy$
So, we get,
\[\begin{align}
  & x.{{e}^{\dfrac{-1}{y}}}=\int{\dfrac{1}{{{y}^{3}}}.{{e}^{\dfrac{-1}{y}}}dy}+C \\
 & x.{{e}^{\dfrac{-1}{y}}}=\int{-t.{{e}^{t}}dt}+C \\
 & x.{{e}^{\dfrac{-1}{y}}}=-\int{t.{{e}^{t}}dt}+C \\
\end{align}\]
Now, let us consider the formula $\int{f\left( x \right)g\left( x \right)dx=}f\left( x \right)\int{g\left( x \right)dx}-\int{{f}'\left( x \right)\left( \int{g\left( x \right)dx} \right)}dx$.
While applying this formula we need to select the functions $f\left( x \right)$ using a rule called as ILATE rule, that is we need to select the function $f\left( x \right)$in the order Inverse, Logarithm, Algebraic, Trigonometric and Exponential and $g\left( x \right)$ will be the other one.
We have two terms in our integral $t$ and ${{e}^{t}}$, let us the ILATE rule for choosing which of them is $f$ and which is $g$. We can see that algebraic function occurs before exponential function. So, we select $f\left( x \right)$ and $g\left( x \right)$ as $f\left( t \right)=t$ and $g\left( t \right)={{e}^{t}}$.
By using the above formula, we get,
\[\begin{align}
  & x.{{e}^{\dfrac{-1}{y}}}=-\int{t.{{e}^{t}}dt}+C \\
 & x.{{e}^{\dfrac{-1}{y}}}=-\left[ t\int{{{e}^{t}}dt}-\int{\left( \dfrac{d}{dt}t \right)\left( \int{{{e}^{t}}dt} \right)dt} \right]+C \\
 & x.{{e}^{\dfrac{-1}{y}}}=-t{{e}^{t}}+\int{1.{{e}^{t}}dt}+C \\
 & x.{{e}^{\dfrac{-1}{y}}}={{e}^{t}}\left( 1-t \right)+C \\
\end{align}\]
Now let us convert $t$ into $\dfrac{-1}{y}$. Then we get,
\[\begin{align}
  & x.{{e}^{\dfrac{-1}{y}}}={{e}^{t}}\left( 1-t \right)+C \\
 & x.{{e}^{\dfrac{-1}{y}}}={{e}^{\dfrac{-1}{y}}}\left( 1+\dfrac{1}{y} \right)+C \\
\end{align}\]
Now, let us substitute $x=1$ and $y=1$ in the equation (1), we get,
\[\begin{align}
  & x.{{e}^{\dfrac{-1}{y}}}={{e}^{\dfrac{-1}{y}}}\left( 1+\dfrac{1}{y} \right)+C \\
 & 1.{{e}^{-1}}=2{{e}^{-1}}+C \\
 & C=2{{e}^{-1}}-{{e}^{-1}} \\
 & C={{e}^{-1}} \\
 & C=\dfrac{1}{e} \\
\end{align}\]
So, we get the equation as \[x.{{e}^{\dfrac{-1}{y}}}={{e}^{\dfrac{-1}{y}}}\left( 1+\dfrac{1}{y} \right)+\dfrac{1}{e}\]
Now, let us substitute $y=2$ in the equation (1), we get,
\[\begin{align}
  & x.{{e}^{\dfrac{-1}{2}}}={{e}^{\dfrac{-1}{2}}}\left( 1+\dfrac{1}{2} \right)+\dfrac{1}{e} \\
 & x.{{e}^{\dfrac{-1}{2}}}={{e}^{\dfrac{-1}{2}}}\left( \dfrac{3}{2} \right)+\dfrac{1}{e} \\
 & x=\dfrac{{{e}^{\dfrac{-1}{2}}}\left( \dfrac{3}{2} \right)+{{e}^{-1}}}{{{e}^{\dfrac{-1}{2}}}} \\
 & x=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}} \\
\end{align}\]

Hence, the answer is \[x=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}\]

Note: The possibilities for making mistakes in this type of problems are, one may make a mistake by considering the formula of integral of product of two functions $f\left( x \right)$ and $g\left( x \right)$ as $\int{f\left( x \right)g\left( x \right)dx=}\left( \int{f\left( x \right)dx} \right)\left( \int{g\left( x \right)dx} \right)$.