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

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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.

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