Question

The D.E whose solution is $y = a{x^2} + bx$ :

Answer 1

Before we get into the problem, we need to know some differentiation formulae.
$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$, using this formula we need to find the first and second derivative of the given solution.
From that we will find the values of a and b.

Complete step-by-step solution:
It is given that the solution of the differential equation is, $y = a{x^2} + bx$.
Let us find the first and second derivatives of $y$.
We know that, $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$.
Therefore, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(a{x^2} + bx)$
On differentiating this with respect to $x$ we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = 2ax + b$
Let us rename it as, ${y_1} = 2ax + b$.
Now let us find the second derivative of $y$.
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {2ax + b} \right)$
On differentiating the above expression with respect to x we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2a$
Let us rename it as, ${y_2} = 2a$.
From this we get, $a = \dfrac{{{y_2}}}{2}$. And from ${y_1}$we get, $b = {y_1} - {y_2}x$.
Thus, we got the vales of a and b. Now let us substitute these in the given solution we get,
$y = \dfrac{{{y_2}}}{2}{x^2} + ({y_1} - {y_2}x)x$
On simplifying this we get,
$\Rightarrow y = \dfrac{{{y_2}}}{2}{x^2} + {y_1}x - {y_2}{x^2}$
On further simplification we get
$\Rightarrow y = \dfrac{{{y_2}{x^2} + 2{y_1}x - 2{y_2}{x^2}}}{2}$
Lets simplify it further.
$\Rightarrow y = \dfrac{{2{y_1}x - {y_2}{x^2}}}{2}$
Taking the $2$to the other side we get,
$\Rightarrow 2y = 2{y_1}x - {y_2}{x^2}$
Let us get all the terms to one side.
$\Rightarrow 2y - 2{y_1}x + {y_2}{x^2} = 0$
Rearranging the terms, we get
$\Rightarrow {y_2}{x^2} - 2{y_1}x + 2y = 0$
Thus, this is the required differential equation whose solution is $y = a{x^2} + bx$.

Note: When the solution is given, we first try to find the values of a and b. This can be done by reducing the given solution by finding its derivative since a and b are the constant values. After finding the values of a and b in terms of derivatives of y, we will substitute it in the solution to find the required differential equation.