Question

Show that \[y=\dfrac{2}{3}{{e}^{x}}+{{e}^{-2x}}\] is a solution of the differential equation \[{y}'+2y=2{{e}^{x}}\]?

Answer 1

Hint: In this question, we have to show something and showing something is a solution to an equation means plugging in the given expression. Therefore, in order to find a solution to this problem, we have to differentiate the $y$ equation and then plug the expression of $y$ and ${y}'$ to that differential equation, and then we have to make sure it holds true.

Complete step-by-step solution:
Here, we are given the differential equation \[{y}'+2y=2{{e}^{x}}\], and we are asked to verify that it works when \[y=\dfrac{2}{3}{{e}^{x}}+{{e}^{-2x}}\].
Since the differential equation needs ${y}'$ as well as $y$, we first have to find $\dfrac{dy}{dx}$ for the given $y$,
We have our equation as:
\[\Rightarrow y=\dfrac{2}{3}{{e}^{x}}+{{e}^{-2x}}\]
On differentiating with respect to $x$, we get our above equation as:
\[\Rightarrow {y}'=\dfrac{dy}{dx}\left( \dfrac{2}{3}{{e}^{x}}+{{e}^{-2x}} \right)\]
On differentiating, we get:
\[\Rightarrow {y}'=\dfrac{2}{3}{{e}^{x}}-2{{e}^{-2x}}\]
Now, as we have expressions for both $y$ and ${y}'$, which we can now plug into our differential equation $\Rightarrow \left( {y}'+2y=2{{e}^{x}} \right)$.
And if this produces a true statement that is equal to our differential equation, then the given expression for $y$ is a valid solution.
Therefore, on substituting the expression for $y$ and ${y}'$ in our differential equation that is \[{y}'+2y=2{{e}^{x}}\], we get:
\[\Rightarrow \left( \dfrac{2}{3}{{e}^{x}}-2{{e}^{-2x}} \right)+2\left( \dfrac{2}{3}{{e}^{x}}+{{e}^{-2x}} \right)=2{{e}^{x}}\]
On simplifying, we get:
\[\Rightarrow \left( \dfrac{2}{3}{{e}^{x}}-2{{e}^{-2x}} \right)+\left( \dfrac{4}{3}{{e}^{x}}+2{{e}^{-2x}} \right)=2{{e}^{x}}\]
On expanding our equation, we get:
\[\Rightarrow \dfrac{2}{3}{{e}^{x}}-2{{e}^{-2x}}+\dfrac{4}{3}{{e}^{x}}+2{{e}^{-2x}}=2{{e}^{x}}\]
On simplifying and cancelling $2{{e}^{-2x}}$ term, we get:
\[\Rightarrow \dfrac{2}{3}{{e}^{x}}+\dfrac{4}{3}{{e}^{x}}=2{{e}^{x}}\]
On solving left-hand side and simplifying, we get:
\[\Rightarrow \left( \dfrac{2}{3}+\dfrac{4}{3} \right){{e}^{x}}=2{{e}^{x}}\]
On simplifying, we get:
\[\Rightarrow \left( \dfrac{6}{3} \right){{e}^{x}}=2{{e}^{x}}\]
On simplifying, we get:
\[\Rightarrow 2{{e}^{x}}=2{{e}^{x}}\]
That is equal to the right-hand side.
Since the differential equation holds true for the given y. Therefore, it is a valid solution of the differential equation.

Note: While solving this type of differential equation, there are two other methods to solve this problem, they could be done using the factor that it is a linear DE with constant coefficients. It can also be solved by using an Integrating Factor but from First Principles. We have to decide which method is efficient and use that method to solve the problem.