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

Last updated date: 19th Jul 2024
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Hint: 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.