Question

$y=\dfrac{x}{x+1}$ is a solution of the differential equation.
A. $y^{2}\dfrac{\text{d}y}{\text{d}x}=x^{2}$
B. $x^{2}\dfrac{\text{d}y}{\text{d}x}=y^{2}$
C. $y\dfrac{\text{d}y}{\text{d}x}=x$
D. $x\dfrac{\text{d}y}{\text{d}x}=y$

Answer 1

Hint: Take the reciprocal on both sides of the given solution for simplicity so that it won’t make the equation complicated while differentiating and also the equation will be simplified such that the division formula for differentiation can be applied during solving part. Differentiate both sides of the equation with respect to x and further solve it to get the required differential equation.




Formula Used:$\dfrac{\text{d}}{\text{d}x}\lgroup\dfrac{u}{v}\rgroup=\dfrac{v\dfrac{\text{d}}{\text{d}x}\lgroup~u\rgroup-u\dfrac{\text{d}}{\text{d}x}\lgroup~v\rgroup}{v^{2}}$



Complete step by step solution:We have already given the solution to the differential equation that needs to be determined, i.e,
$y=\dfrac{x}{x+1}$
Let us take the reciprocal on both sides of the above equation so that by further solving it can be written in simplest form.
$\dfrac{1}{y}=\dfrac{x+1}{x}$
Now,
$\dfrac{1}{y}=\dfrac{x}{x}+\dfrac{1}{x}$
$\Rightarrow\dfrac{1}{y}=1+\dfrac{1}{x}$
Differentiate both sides with respect to x,
$\dfrac{\text{d}}{\text{d}x}\lgroup\dfrac{1}{y}\rgroup=\dfrac{\text{d}}{\text{d}x}\lgroup1+\dfrac{1}{x}\rgroup$
The division rule of differentiation gives the following formula,
$\dfrac{\text{d}}{\text{d}x}\lgroup\dfrac{u}{v}\rgroup=\dfrac{v\dfrac{\text{d}}{\text{d}x}\lgroup~u\rgroup-u\dfrac{\text{d}}{\text{d}x}\lgroup~v\rgroup}{v^{2}}$
In the LHS case, let u=1 and v=y whereas in the case of RHS let u=1 and v=x. Also, the differentiation of the constant is always 0.
Now, let's apply all these in the equation that needs to be differentiated.
$\lgroup\dfrac{-1}{y^{2}}\rgroup\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}}{\text{d}x}\lgroup1\rgroup+\dfrac{\text{d}}{\text{d}x}\lgroup\dfrac{1}{x}\rgroup$
Further solving,
$\lgroup\dfrac{-1}{y^{2}}\rgroup\dfrac{\text{d}y}{\text{d}x}=0-\lgroup\dfrac{1}{x^{2}}\rgroup$
$x^{2}\dfrac{\text{d}y}{\text{d}x}=y^{2}$



Option ‘B’ is correct

Note: We are given a solution, which we plug into the differential equation and check to see if the resulting equation is correct. In this way, we can cross-verify our resultant solution.