Question

Number of straight lines which satisfy the differential equation \[\dfrac{dy}{dx}+x{{\left( \dfrac{dy}{dx} \right)}^{2}}-y=0\] is :
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: Given the differential equation \[\dfrac{dy}{dx}+x{{\left( \dfrac{dy}{dx} \right)}^{2}}-y=0\] is, we have to find out the number of straight lines which satisfy the given differential equation. Consider a line equation then differentiate it, find the slope and substitute the value in the differential equation and solve for roots.
Complete step-by-step answer:
Now considering the line \[y=mx+c\]
differentiate the line equation with respect to x, then we get the equation as \[\dfrac{dy}{dx}=m\]
This equation satisfies the given differential equation \[\dfrac{dy}{dx}+x{{\left( \dfrac{dy}{dx} \right)}^{2}}-y=0\]
So substituting \[\dfrac{dy}{dx}=m\] and \[y=mx+c\] in the above equation gives the equation as \[m+x{{m}^{2}}-mx-c=0\]
Now writing the equation with common terms we get the equation as \[x({{m}^{2}}-m)+(m-c)=0\]
In the above equation to satisfy both the terms should become zero.
The coefficient of x term should become zero and the constant term should also become zero.
Now equating the coefficient of x term to zero we get the equation as \[({{m}^{2}}-m)=0\]
Now taking m as common we get the further equation as \[m(m-1)=0\]
From the above equation we can conclude that m can be zero or m can be 1
\[\Rightarrow \] \[\begin{align}
  & m=0 \\
 & m=1 \\
\end{align}\]
Now equating the constant term of the equation to zero, we get the further equation as \[(m-c)=0\]
From the above equation we can conclude that m is equal to c
\[\Rightarrow m=c\]
Here the conclusion is m can take either 0 or 1 and m=c.
So two straight lines satisfy the above differential equation.
Note: Here in the given question students may forget that m=c should not be considered as a different equation because both conditions are necessary for being a straight line equation.