Question

Find the general solution for the following differential equation \[\dfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}\].

Answer 1

Hint: A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. A solution of a differential equation is a relation between the variables , is free of derivatives of any order, and which satisfies the differential equation identically.

Complete step-by-step answer:
Steps for solving this kind of differential equations are:
Put the given differential equation in the form \[\dfrac{{dy}}{{dx}} + Py = Q\].
Find the values for P and Q by comparing with the general equation.
Find the integrating factor as I.F. \[ = {e^{\int {Pdx} }}\].
Solution of the given differential equation will be \[y \times I.F. = \int {Q \times I.F.dx} + C\].
There are various types of differential equations; such as – homogeneous and non-homogeneous, linear and nonlinear, ordinary and partial. The differential equation may be of the first order, second order and even more than that.
An integrating factor is a function used to solve differential equations. It is a function in which an ordinary differential equation can be multiplied to make the function integrable.
We are given the differential equation \[\dfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}\].
Comparing this equation with the general equation \[\dfrac{{dy}}{{dx}} + Py = Q\] we get
\[P = 3\] and \[Q = {e^{ - 2x}}\]
Therefore we get integration factor \[I = {e^{\int {3dx} }}\]\[ = {e^{3x}}\]
Now solution of the given equation is
\[y \times {e^{3x}} = \int {{e^{3x}}.} {e^{ - 2x}}dx\]
On simplification we get ,
\[y \times {e^{3x}} = \int {{e^x}.} dx\]
On solving the integral we get ,
\[y \times {e^{3x}} = {e^x} + C\]
Where C is the integration constant.
Therefore we get ,
\[y = {e^{ - 2x}} + C{e^{ - 3x}}\]
Therefore, the required solution is \[y = {e^{ - 2x}} + C{e^{ - 3x}}\].

Note: In Mathematics, Differentiation can be defined as a derivative of a function with respect to an independent variable. In Calculus, Differentiation can be applied to measure the function per unit change in the independent variable.

And the degree of the differential equation is the power of the highest order derivative, where the original equation is represented in the form of a polynomial equation in derivatives such as \[{y}',{y}'',\text{ }{y}'''\], and so on.