Question

Solve $\dfrac{{dy}}{{dx}} + 2y = \sin x$

Answer 1

In the question we are given a linear differential equation. To understand this question properly we will first understand the meaning of linear differential equation and method to solve it. Then we will find the integrating factor of our given equation and using that we will find the general solution of our given differential equation.

Complete step by step solution:
Linear Differential Equation:
Linear Differential Equation is the differential equation in which order and degree of differential equation is one.
General form of linear differential equation is represented as
$\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$ Where $P$ and $Q$ are functions of $x$
For solving linear differential equation
First we calculate integrating factor of the equation
Formula for integrating factor $(I.F)$ is
$I.F = {e^{\int_{}^{} {Pdx} }}$
Then using this we calculate general solution of linear differential equation.
$y \times I.F = \int_{}^{} {Q \times I.Fdx}$
Remember in this linear differential equation $x$ is independent variable and $y$ is dependent variable.
Now let us proceed to our equation.
In the question we are given a linear differential equation.
General form of a linear differential equation is
$\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$ Where $P$ and $Q$ are functions of $x$
Our given equation is
$\dfrac{{dy}}{{dx}} + 2y = \sin x$
Comparing it with general equation we get
$P = 2 \\\ Q = \sin x \\\$
Now we will calculate Integrating Factor $(I.F)$
Formula for integrating factor is
$I.F = {e^{\int_{}^{} {Pdx} }}$
According to our given equation
$P = 2$
Substituting value of $P$ .
$I.F = {e^{\int_{}^{} {2dx} }}$
As we know $\int_{}^{} {adx = ax + c}$ where $a$ is any constant and $c$ is an integrating constant.
Here we will not consider $c$ in integrating factor
So
$I.F = {e^{2x}}$
Now we will find general solution of our given equation
General solution of a linear differential equation is given as
$y \times I.F = \int_{}^{} {Q \times I.Fdx}$
Substituting the values of $I.F$ and $Q$
$I.F = {e^{2x}}$
$Q = \sin x$
General solution is
$y \times {e^{2x}} = \int_{}^{} {\sin x \times {e^{2x}}dx}$
We can write it as
$y{e^{2x}} = \int_{}^{} {\sin x{e^{2x}}dx}$
$\Rightarrow y{e^{2x}} = \int_{}^{} {{e^{2x}}\sin xdx}$
We can use a direct formula of integration here
$\int_{}^{} {{e^{ax}}\sin bxdx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}(a\sin bx - b\cos bx) + C}$
Comparing it with the integral we have to find
$a = 2 \\\ b = 1 \\\$
Now we will substitute this in the formula
$\int_{}^{} {{e^{2x}}\sin 1xdx = \dfrac{{{e^{2x}}}}{{{2^2} + {1^2}}}(2\sin 1x - 1\cos 1x) + C}$
We can write it as
$\int_{}^{} {{e^{2x}}\sin xdx = \dfrac{{{e^{2x}}}}{{{2^2} + {1^2}}}(2\sin x - \cos x) + C}$
Simplifying this
$\int_{}^{} {{e^{2x}}\sin xdx = \dfrac{{{e^{2x}}}}{{4 + 1}}(2\sin x - \cos x) + C} \\\ \Rightarrow \int_{}^{} {{e^{2x}}\sin xdx = \dfrac{{{e^{2x}}}}{5}(2\sin x - \cos x) + C} \\\$
Now we will substitute this in our general solution.
$y{e^{2x}} = \int_{}^{} {{e^{2x}}\sin xdx} \\\ \Rightarrow y{e^{2x}} = \dfrac{{{e^{2x}}}}{5}(2\sin x - \cos x) + C \\\$
$\therefore$ General solution of our given linear differential equation $\dfrac{{dy}}{{dx}} + 2y = \sin x$ is $y{e^{2x}} = \dfrac{{{e^{2x}}}}{5}(2\sin x - \cos x) + C$ , where $C$ is an integrating constant.

Note: A differential equation is an equation which involves derivatives. To solve differential equations of first order and first degree we have four main methods named as variable separable, reducible to variable separable, homogeneous equation method and solving by using integrating factor.