Question

What is the solution to the differential equation $\left( \dfrac{dy}{dx} \right)-y-{{e}^{3x}}=0$?

Answer 1

In this problem, we have to find the solution for the given linear differential equation. We should know that, the differential equation of the form $\dfrac{dy}{dx}+Py=Q$ which is linear in y, has a solution $y{{e}^{\int{Pdx}}}=\int{Q}\times {{e}^{\int{Pdx}}}dx+C$ where, ${{e}^{\int{Pdx}}}$ is the integrating factor. Here we can first find the integrating factor and we can substitute in the solution equation to get the final answer.

Complete step-by-step solution:
We know that the given linear differential equation is,
$\left( \dfrac{dy}{dx} \right)-y-{{e}^{3x}}=0$
We can write it as,
$\Rightarrow \left( \dfrac{dy}{dx} \right)-y={{e}^{3x}}$……. (1)
We can see that the given linear differential equation is of the form,
$\dfrac{dy}{dx}+Py=Q$
Where, $P=-1,Q={{e}^{3x}}$.
We know that the differential equation of the form $\dfrac{dy}{dx}+Py=Q$ which is linear in y, has a solution $y{{e}^{\int{Pdx}}}=\int{Q}\times {{e}^{\int{Pdx}}}dx+C$……. (2)
where, ${{e}^{\int{Pdx}}}$ is the integrating factor.
We can now find the integrating factor, we get
$\Rightarrow {{e}^{\int{-1dx}}}={{e}^{-x}}$
We can now substitute the above integrating factor and the value of Q in the solution equation (2), we get
$\Rightarrow y{{e}^{-x}}=\int{{{e}^{3x}}}\times {{e}^{-x}}dx+C$
We can now simplify the above step, we get
$\Rightarrow y{{e}^{-x}}=\int{{{e}^{2x}}}dx+C$
We can now differentiate the right-hand side, we get
$\Rightarrow y{{e}^{-x}}=\dfrac{1}{2}{{e}^{2x}}+C$
We can now take the exponential part from the left-hand side to the right-hand side, we get
$\Rightarrow y=\dfrac{1}{2}{{e}^{3x}}+C{{e}^{x}}$
Therefore, the solution of the given differential equation $\left( \dfrac{dy}{dx} \right)-y-{{e}^{3x}}=0$ is $y=\dfrac{1}{2}{{e}^{3x}}+C{{e}^{x}}$

Note: We should always remember that a linear differential equation of the form $\dfrac{dy}{dx}+Py=Q$, can be differentiated using the formula $y{{e}^{\int{Pdx}}}=\int{Q}\times {{e}^{\int{Pdx}}}dx+C$, where ${{e}^{\int{Pdx}}}$ is the integrating factor. We should also know some of the exponential formulas to simplify the steps for the final answer.