Question

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

Answer 1

Hint: 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.