Question

The solution of differential equation, $\dfrac{dy}{dx}={{\left( y-x \right)}^{2}}$ when $y\left( 1 \right)=1$, is –
(a) ${{\log }_{e}}\left| \dfrac{2-y}{2-x} \right|=2\left( y-1 \right)$
(b) $-{{\log }_{e}}\left| \dfrac{1+x-y}{1-x+x} \right|=x+y-2$
(c) ${{\log }_{e}}\left| \dfrac{2-x}{2-y} \right|=x-y$
(d) $-{{\log }_{e}}\left| \dfrac{1-x+y}{1+x-y} \right|=2\left( x-1 \right)$

Answer 1

Hint: Here, we will use a substitution method to find the solution of the given differential equation. After finding the general solution, we will use the given condition $y\left( 1 \right)=1$ to find the value of constant obtained in the general solution and thus conclude our final answer. We will use the basic differentiation and integration for solving the differentiation equation. Integration formula that we will use are given as $\int{\dfrac{dx}{1-{{x}^{2}}}=\dfrac{1}{2}}{{\log }_{2}}\dfrac{1+x}{1-x}$ and $\int{1dx=x}$

Complete step-by-step solution
Given equation is $\dfrac{dy}{dx}={{\left( y-x \right)}^{2}}$ with given condition $y\left( 1 \right)=1$, which means when we put value of $x$ as $1$, the solution value of $y$ is also $1$.
$\dfrac{dy}{dx}={{\left( y-x \right)}^{2}}~~~~~~~~...\left( 1 \right)$
We will use this substitution method to solve this equation, thus we will put $t$ in place of $\left( x-y \right)$. Let $\left( x-y \right)=t$.
Let us take derivative with respect to $x$ on both sides, we get –
$\begin{align}
 & \dfrac{dx}{dx}-\dfrac{dy}{dx}=\dfrac{dt}{dx} \\
 &\Rightarrow \dfrac{dy}{dx}=1-\dfrac{dt}{dx} \\
\end{align}$
Putting both values in equation (1), we get –
\[\dfrac{dt}{dx}=1-{{t}^{2}}\]
By using a variable separable method, let us take terms containing $t$ on one side and terms containing $x$ on the other side. We will get –
\[\dfrac{dt}{1-{{t}^{2}}}=dx\]
To solve this equation, let us take integration on both sides. We get –
$\int{\dfrac{1}{1-{{t}^{2}}}.dt=\int{1.}dx}$
Solving LHS of the equation first, we get –
\[\int{\dfrac{1}{1-{{t}^{2}}}.dt}\]
Factoring numerator and denominator , we get –
$\int{\dfrac{1}{\left( 1-t \right)\left( 1+t \right)}.dt}$
Using partial fraction decomposition,
\[\begin{align}
&\dfrac{1}{\left( 1-t \right)\left( 1+t \right)}=\dfrac{{{A}_{1}}}{\left( 1+t \right)}+\dfrac{{{A}_{2}}}{\left( 1-t \right)} \\
&\Rightarrow \dfrac{1}{\left( 1-t \right)\left( 1+t \right)}=\dfrac{{{A}_{1}}\left( 1-t \right)+{{A}_{2}}\left( 1+t \right)}{\left( 1+t \right)} \\
 &\Rightarrow 1={{A}_{1}}\left( 1-t \right)+{{A}_{2}}\left( 1+t \right) \\
\end{align}\]
Putting $t=1$, \[\begin{align}
  & {{A}_{2}}\left( 1+1 \right)=1 \\
 &\Rightarrow {{A}_{2}}=\dfrac{1}{2} \\
\end{align}\]
Putting $t=-1$, \[\begin{align}
  & {{A}_{1}}\left( 1+1 \right)=1 \\
 &\Rightarrow {{A}_{1}}=\dfrac{1}{2} \\
\end{align}\]
Therefore,
$\begin{align}
  & \int{\dfrac{1}{\left( 1-t \right)\left( 1+t \right)}.dt=}\int{\dfrac{1}{2\left( 1+t \right)}.dt+\int{\dfrac{1}{2\left( 1-t \right)}}}.dt \\
 &\Rightarrow \int{\dfrac{1}{\left( 1-t \right)\left( 1+t \right)}.dt=}\dfrac{1}{2}\left[ \int{\dfrac{1}{\left( 1+t \right)}.dt+\int{\dfrac{1}{\left( 1-t \right)}}}.dt \right] \\
\end{align}$
Let $\int{\dfrac{1}{\left( 1+t \right)}.dt={{I}_{1}}}$ and $\int{\dfrac{1}{\left( 1-t \right)}.dt={{I}_{2}}}$
Let us solve for \[{{I}_{1}}\] and \[{{I}_{2}}\] separately.
\[{{I}_{1}}=\int{\dfrac{dt}{\left( 1+t \right)}}\]
Let $1+t=u$
$\begin{align}
  & dt=du \\
 &\Rightarrow {{I}_{1}}=\int{\dfrac{du}{u}} \\
\end{align}$
Since $\int{\dfrac{dx}{x}={{\log }_{e}}\left| x \right|}$, therefore –
${{I}_{1}}=\int{\dfrac{du}{u}}={{\log }_{e}}\left| u \right|$, where $u=1+t$
Therefore, ${{I}_{1}}=\ln \left| 1+t \right|$
\[{{I}_{2}}=\int{\dfrac{dt}{\left( 1-t \right)}}\]
Let $1-t=v$
$\begin{align}
  & -dt=dv \\
 \Rightarrow {{I}_{2}}=\int{\dfrac{-dv}{v}}=-\int{\dfrac{dv}{v}} \\
\end{align}$
Since, $\int{\dfrac{dx}{x}={{\log }_{e}}\left| x \right|}$, therefore –
${{I}_{2}}=-\int{\dfrac{dv}{v}}=-{{\log }_{e}}\left| v \right|$, where $v=1-t$
Therefore,
 Hence, \[\int{\dfrac{dt}{1-{{t}^{2}}}}\] becomes $\dfrac{1}{2}\left( {{I}_{1}}+{{I}_{2}} \right)=\dfrac{1}{2}\left( {{\log }_{e}}\left| 1+t \right|-{{\log }_{e}}\left| 1-t \right| \right)$
Since, ${{\log }_{e}}x-{{\log }_{e}}y={{\log }_{e}}\left( \dfrac{x}{y} \right)$
Therefore, \[\int{\dfrac{dt}{1-{{t}^{2}}}}=\dfrac{1}{2}\left[ {{\log }_{e}}\left| \dfrac{1+t}{1-t} \right| \right]=\dfrac{1}{2}{{\log }_{e}}\left| \dfrac{1+t}{1-t} \right|\]
Also, we know that $\int{1dx=x}$. So using both results in above equation, we get –
\[\dfrac{1}{2}{{\log }_{e}}\left| \dfrac{1+t}{1-t} \right|=x+c\], where $c$ is any constant.
Now the solution is obtained in terms of $t$, but we have to find it in terms of $x$ and $y$. Thus putting value of $x-y=t$ back, we get –
\[\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1+x-y}{1-x+y} \right)=x+c~~~~~~~~...\left( 2 \right)\]
Hence, we have obtained a general solution of the given differential equation. Now let us use the given condition $y\left( 1 \right)=1$to find the value of $c$ which will help us in finding the required answer.
Putting $x=1$ and $y=1$ in equation (2), we get –
\[\begin{align}
  & \dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1+1-1}{1-1+1} \right)=1+c \\
 &\Rightarrow \dfrac{1}{2}{{\log }_{e}}\left( 1 \right)=1+c \\
\end{align}\]
As we know ${{\log }_{e}}\left( 1 \right)=0$, thus \[\begin{align}
  & 1+c=0 \\
 & \Rightarrow c=-1 \\
\end{align}\]
Hence, we have obtained a value of $c$ which is $-1$. Now putting this value in equation (2), we get –
\[\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1+x-y}{1-x+y} \right)=x-1\]
Multiplying both sides by 2, we get –
\[{{\log }_{e}}\left( \dfrac{1+x-y}{1-x+y} \right)=2\left( x-1 \right)\]
As we know that \[{{\log }_{e}}\left( x \right)=-{{\log }_{e}}\dfrac{1}{x}\] . Thus we get,
\[-{{\log }_{e}}\left( \dfrac{1-x+y}{1+x-y} \right)=2\left( x-1 \right)\], which is our required answer.
Hence, the correct option is (d).

Note: For solving differential equations, students should practice more and more to identify which method should be used. Also, students should not get confused given condition, the value inside the bracket represents the value of $x$ and the value at RHS represents the value of $y$. Students should learn integration formulas as there will be no time for deriving them during the exam. Don’t forget to put back the substituted value, the value we assume should always be eliminated at the final answer.