# Verify that each given function is a solution of the differential equation:

$\boldsymbol{ t y’ \ – \ y \ = \ t^2, \ y \ = \ 3 t \ + \ t^2 }$

The aim of this question is to learn the basic verification procedure for solutions to differential equations.

It’s simply a reverse calculation procedure. You start with the given value of $y$ and then successively differentiate it as per the order of the differential equation. Once you have all the derivatives, we simply put them into the given differential equation to check whether the equation is properly satisfied or not. If the equation is satisfied, the given solution is indeed a root/solution to the given differential equation.

Step (1): Differentiating $y$ with respect to $t$.

Given:

$y \ = \ 3 t \ + \ t^2$

Differentiating:

$y’ \ = 3 \ + \ 2 t \ … \ … \ … \ (1)$

Step (2): Substitute the given values.

Given:

$t y’ \ – \ y \ = \ t^2$

$\Rightarrow t \ ( \ 3 \ + \ 2 t \ ) \ – \ y \ = \ t^2$

$\Rightarrow y’ \ = \ t \ + \ \dfrac{ y }{ t }$

Substituting values of $y’$ and $y$:

$t \ ( \ 3 \ + \ 2 t \ ) \ – \ ( \ 3 t \ + \ t^2 \ ) \ = \ t^2$

$\Rightarrow 3 t \ + \ 2 t^2 \ – \ 3 t \ – \ t^2 \ ) \ = \ t^2$

$\Rightarrow 3 t \ + \ 2 t^2 \ = \ 3 t \ + \ 2 t^2$

Since the equation is satisfied, given solution indeed belongs to the given differential equation.

## Numerical Result

$y \ = \ 3 t \ + \ t^2$ is the solution to the differential equation $t y’ \ – \ y \ = \ t^2$.

## Example

Make sure that each given function is a solution of the differential equation:

$\boldsymbol{ y^{ ” } \ – \ 4 y \ = \ 0, \ y \ = \ e^{ 2 t } }$

Step (1): Differentiating $y$ with respect to $t$.

Given:

$y \ = \ e^{ 2 t }$

Differentiating once:

$y’ \ = \ 2 e^{ 2 t }$

Differentiating again:

$y^{ ” } \ = \ 4 e^{ 2 t }$

Step (2): Substitute the given values.

Given:

$y^{ ” } \ – \ 4 y \ = \ 0$

Substituting values of $y’$ and $y$:

$4 e^{ 2 t } \ – \ 4 ( e^{ 2 t } ) \ = \ 0$

$4 e^{ 2 t } \ = \ 4 ( e^{ 2 t } )$

Since the equation is satisfied, the given solution indeed belongs to the given differential equation.