Question

The curve satisfying the differential equation, $ydx-\left( x+3{{y}^{2}} \right)dy=0$ and passing through the point $\left( 1,1 \right)$ also passes through the point \[\]
A. $\left( \dfrac{1}{4},-\dfrac{1}{2} \right)$\[\]
B. $\left( \dfrac{1}{4},\dfrac{1}{2} \right)$\[\]
C. $\left( -\dfrac{1}{3},\dfrac{1}{3} \right)$\[\]
D. $\left( \dfrac{1}{3},-\dfrac{1}{3} \right)$\[\]

Answer 1

Hint: Convert the linear differential equation to the form ${{y}^{'}}+yP\left( x \right)=Q\left( x \right)$ and solve it by integrating factor. Finally use the given point to put in the solution as initial condition and check each option whether they satisfy or not. \[\]

Complete step by step answer:
We know that the solution of linear differential equation of the form ${{y}^{'}}+yP\left( x \right)=Q\left( x \right)$ is $y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}{{e}^{\int{P\left( x \right)dx}}}dx+C$ where $P\left( x \right)$ and $Q\left( x \right)$ are real differentiable functions, $C$ is a real constant and the integrating factor is $\text{IF}={{e}^{\int{P\left( x \right)dx}}}$.\[\]

The given differential equation of the curve is
\[ydx-\left( x+3{{y}^{2}} \right)dy=0\]
 We begin by separating the differentials and variables at different side ,
\[\begin{align}
  & ydx-\left( x+3{{y}^{2}} \right)dy=0 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x+3{{y}^{2}}} \\
\end{align}\]
 We cannot solve towards $y=f\left( x \right)$ as we cannot convert. So we proceed towards $x=f\left( y \right)$
\[\begin{align}
  & ydx-\left( x+3{{y}^{2}} \right)dy=0 \\
 & \Rightarrow \dfrac{dx}{dy}=\dfrac{x+3{{y}^{2}}}{y} \\
 & \Rightarrow \dfrac{dx}{dy}-\left( \dfrac{1}{y} \right)x=3y \\
\end{align}\]
Which of the form ${{x}^{'}}+xP\left( y \right)=Q\left( y \right)$ where $P\left( y \right)=\dfrac{-1}{y},Q\left( y \right)=3y$ The solution of the differential equation in this form is \[x{{e}^{\int{P}\left( y \right)dy}}=\int{Q\left( y \right)}{{e}^{\int{P}\left( y \right)dy}}dy+C\] . We first find the integrating factor,
\[{{e}^{\int{P\left( y \right)dy}}}={{e}^{\int{\dfrac{-1}{y}dy}}}={{e}^{-\ln y}}={{e}^{\ln \dfrac{1}{y}}}=\dfrac{1}{y}(\text{using }\int{\dfrac{1}{y}dy=\ln y\text{ and }\ln {{e}^{\ln a}}=a}\text{ })\]
Using it in the standard solution ,
\[\begin{align}
  & x\left( \dfrac{1}{y} \right)=\int{3y}\left( \dfrac{1}{y} \right)dy \\
 & \Rightarrow \dfrac{x}{y}=3y+C \\
 & \Rightarrow x=3{{y}^{2}}+Cy \\
\end{align}\]
Alternatively for faster solution we use separation of variables,
\[\begin{align}
  & ydx-\left( x+3{{y}^{2}} \right)dy=0 \\
 & \Rightarrow ydx-xdy=3{{y}^{2}}dy \\
 & \Rightarrow \dfrac{ydx-xdy}{{{y}^{2}}}=\dfrac{3{{y}^{2}}dy}{{{y}^{2}}} \\
\end{align}\]
We know from differential calculus that $d\left( \dfrac{x}{y} \right)=\dfrac{ydx-xdy}{{{y}^{2}}}$. We use this fact while integrating bothside,
\[\begin{align}
  & \int{\dfrac{ydx-xdy}{{{y}^{2}}}}=\int{3dy} \\
 & \Rightarrow \dfrac{x}{y}=3y+C \\
\end{align}\]

In both methods we get the same result. Now we put the point $\left( 1,1 \right)$ in above equation,
\[\begin{align}
  & x=3{{y}^{2}}+Cy \\
 & \Rightarrow 1=3{{\left( 1 \right)}^{2}}+C \\
 & \Rightarrow C=-2 \\
\end{align}\]
So the solution equation of the given differential equation is $x=3{{y}^{2}}-2y$. \[\]
Checking option A $\left( \dfrac{1}{4},\dfrac{1}{2} \right),3{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{1}{2} \right)\ne \dfrac{1}{4}$, \[\]
Checking option B $\left( \dfrac{1}{4},\dfrac{1}{2} \right),3{{\left( \dfrac{-1}{2} \right)}^{2}}-2\left( \dfrac{-1}{2} \right)\ne \dfrac{1}{4}$\[\]
Checking option C $\left( \dfrac{1}{3},-\dfrac{1}{3} \right),3{{\left( \dfrac{-1}{3} \right)}^{2}}-2\left( \dfrac{-1}{3} \right)=\dfrac{1}{3}$.
Checking option D $\left( -\dfrac{1}{3},-\dfrac{1}{3} \right),3{{\left( \dfrac{1}{3} \right)}^{2}}-2\left( \dfrac{1}{3} \right)\ne \dfrac{1}{3}$\[\]

So, the correct answer is “Option C”.

Note: The question tests your knowledge of linear differential equations with solutions involving integrating factors. Careful solving of simultaneous equation substitution and usage of formula will lead us to arrive at the correct result. Similarly a question could be made what type of curve the solution represents. We can find from the general equation second degree that the curve is a rightward parabola.