
The D.E whose solution is \[y = a{x^2} + bx\] :
Answer
412.8k+ views
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.
\[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\], 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.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
