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

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**.

## Expert Answer

**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.