Question

Find general solution of differential equation $\dfrac{{dy}}{{dx}} + y = 1\left( {y \ne 1} \right).$

Answer 1

Hint: Given equation in the question is based on the type of solutions of the differential concept.
This type of equation is solved by one of the method,
$ \Rightarrow \dfrac{{dy}}{{dx}} + Py = Q$_ _ _ _ _ _ _ _ _ _ $\left( 1 \right)$
And integrated factor concept,
$ \Rightarrow {e^{\int {Pdx} }}.$

Complete step-by-step solution:
The given equation in the question$\dfrac{{dy}}{{dx}} + y = 1\left( {y \ne 1} \right).$ is first order linear equation.
Comparing this equation with$\left( 1 \right)$,
We get,
$ \Rightarrow P = 1,Q = 1$
So, now integrated factor,
I.F. $ \Rightarrow {e^{\int {Pdx} }}.$
$ \Rightarrow {e^{\int {1dx} }} = {e^x}$
Hence, the solution can be given by,
$ \Rightarrow y \times I.F = \int {q \times I.F.} dx$
$ \Rightarrow y{e^x} = \int {{e^x}dx} \\
\Rightarrow y{e^x} = e^x + C \\
 \Rightarrow y = 1 + C{e^{ - x}}. \\ $
Therefore $1 + C{e^{ - x}}$is the solution of the differential equation.

Note: $ \Rightarrow $ A differential equation is an equation involving an unknown function y=f(x) and one or more of its derivatives.
$ \Rightarrow $ A solution to a differential equation is a function y=f(x) that satisfies the differential equation when f and its derivatives are substituted into the equation.
$ \Rightarrow $When the arbitrary constant of the general solution takes some unique value, then the solution becomes the particular solution of the equation. By using the boundary conditions the particular solution of a differential equation is obtained.