Question

Find one-parameter families of solution curves of the following differential equation:-

\[\frac{dy}{dx} + 3y = e^{mx}\], m is a given real number.

Solve the following differential equation:-

\[\frac{dy}{dx} + 3y = e^{mx}\], m is a given real number.

Answer 1

 We have, 
\[\frac{dy}{dx} + 3y = e^{mx} . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 3 \]
\[Q = e^{mx} \]
\[ \therefore I.F. = e^{\int P\ dx} \]
\[ = e^{\int3 dx} \]
\[ = e^{3x} \]
\[\text{ Multiplying both sides of }(1)\text{ by }e^{3x} ,\text{ we get }\]
\[ e^{3x} \left( \frac{dy}{dx} + 3y \right) = e^{3x} e^{mx} \]
\[ \Rightarrow e^{3x} \frac{dy}{dx} + 3 e^{3x} y = e^\left( m + 3 \right)x \]
Integrating both sides with respect to x, we get
\[y e^{3x} = \int e^\left( m + 3 \right)x dx + C .............\left(\text{when }m + 3 \neq 0 \right) \]
\[ \Rightarrow y e^{3x} = \frac{e^\left( m + 3 \right)x}{m + 3} + C\]
\[ \Rightarrow y = \frac{e^{mx}}{m + 3} + C e^{- 3x} \]
\[y e^{3x} = \int e^{0 \times x} dx + C ...........\left(\text{when }m + 3 = 0 \right) \]
\[ \Rightarrow y e^{3x} = \int dx + C\]
\[ \Rightarrow y e^{3x} = x + C\]
\[ \Rightarrow y = \left( x + C \right) e^{- 3x} \]
Hence, 
\[ y = \frac{e^{mx}}{m + 3} + C e^{- 3x} ,\text{ where }m + 3 \neq 0\]
and
\[y = \left( x + C \right) e^{- 3x} ,\text{ where }m + 3 = 0 \text{ are required solutions.}\]