Solution Of Differential Equations With Perfect Examples
Solution Of Differential Equations:
From ancient Greek Mathematics equations are the short form of a long analytical problem .
A large problem can be solved in a easy way by converting it into a simple equation.
Differential equations are the evolved form of equations .
So here i want to discuss about some problems of Differential equations and it's methods of solving problems.
For example_:
d^2y/dx^2+dy/dx+y=3
Let us consider some problems and solutions regarding this.
Examples :
1. d^2y/dx^2+y=x
2. y"+y'+6y=0
Solutions :
1. Given ,dy/dx+y=x
It is the linear equation of the form dy/dx+p(x)y=q(x)
In this case p(x)=1 and q(x) = x
So to solve this we have to find integrating factor i.e I.F(μ) = e^∫p(x)dx
so μ= e^∫dx= e^x
Multiplying I.F to the above equation we get ,
e^xdy/dx+e^x y=xe^x
=>d/dx(y e^x)=xe^x
=>y e^x=∫e^x x dx
=>y e^x=xe^x - e^x+c
=>y= x - 1 + ce^-x
Which is the required solution.
2. Given,
y"+y'+6y=0
To solve this let us put y= e^mx
y'=dy/dx=me^mx, y"= d^2y/dx^2=m^2e^mx
Putting all these values in the given equation we get,
m^2e^mx+me^mx+6e^mx=0
=>(m^2+m+6)e^mx=0
As
e^mx≠0 so m^2+m+6=0
Which is a quadratic equation so
m=[ -1±(1-4*6)^1/2]/2 = [-1±23^1/2i]/2
So y = e^-x/2(c1cos23^1/2x+c2sin23^1/2x)
Hence it is the required solution.
Here I have some questions related to differential equation .
I want you to solve these equations and answer me all these .
For solving any problem we need to analyse it first , then we can solve it in a easy way . So thats the matter .
Questions :
- dy / dx = 2xe^x² .
- dy / dx = e²ˣ + x .
References :
- Numerical Solution of Partial Differential Equations: Finite Difference Methods , By Oxford Applied Mathematics and Computing Science Series
- Numerical Solution of Differential Equations , By M K Jain
- Applied Partial Differential Equations | Fourth Edition | By Pearson
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