Solving Initial Value Problem

Introduction for solving the initial value problem:

Initial value problem is the part of differential equation in mathematics. Initial value problem is an ordinary differential equation joint with particular value, called the initial condition, of the strange function at a known point in the area of the result. In this article we shall discuss  how to solve these initial value problems in the differential equation Having problem with Integral Calculator with Steps keep reading my upcoming posts, i will try to help you.

Example Problems on Solving the Initial Value Problem:

Example 1:

Solving the differential equation `(dy/dx)=4y` using initial condition `y(1)=2`

Solution:

Given differential equation is `(dy/dx)=4y` , initial condition is `y(1)=2`

Step 1:

4y = 0 when y = 0 Thus the constant y(x) = 0 is a particular solution.

Step 2:

Separate the variables on both sides,

`(dy)/y=4 dx`

Step 3:

Take the integral on both sides,

`ln y=4x+B`

step 4:

Take ln on both sides,

`y=e^B e^4x`

`y=Ce^4x`

where `C = eB` if y > 0, and `C = -eB` if y < 0.

Step 5:

General solution: `y(x)=Ce^(4x)` where C is any constant (C can be 0 from Step 1).

Step 6:

Substitute 1 for x and 2 for y

`2=Ce^4,`

`C=2e^(-4)`

The final solution is `y(x)=2e^(-4) e^(4x) =2e^(-(4-4x))`

Example 2:

Solve the initial value problem

Solution:

`y^'' -3y^' +2y=0` , `y(0)=1` ,  `y^' (0)=-1`

First we find the general solution to the differential equation. We can find the general solution to the homogeneous equation by guessing a solution in the form `y(t)=e^(rt)`

`y^'' -3y^' +2y=0`

`(r^2 -3r +2) e^(rt)`

`(r-1)(r-2) e^(rt)`

if we choose r = 1,2 then we find the general solution `y_c (t)=C_1 e^(-t)+C_2 e^(-2t)` for the homogeneous equation.

Now we find the solution corresponding to the initial conditions. We have the derivatives

`y_c (t)=C_1 e^(-t)+C_2 e^(-2t)`

`y(0)=C_1+C_2`

`y^'(t)=C_1 e^(-t)+C_2 e^(-2t)`

`y^'(0)=C_1-C_2`

We see that `C_1+C_2=1 and C_1-C_2=-1`

Solving the above equations we get `C_1=0` and `C_2=1`

Substitute these values we get the solution

`y^' (t)=e^(-t)`


I have recently faced lot of problem while learning how to multiply mixed numbers by fractions, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problems on Solving Initial Value Problems:

Problem 1:

Solve the initial value problem

`y^'' +3y^' +2=0` , `y(0)=1` , `y^' (0)=-1`

Answer: `y^(t)=e^(-t)`

Problem 2:

Solve the initial value problem

`y^'' +7y^' +12=0` , `y(0)=1` , `y^' (0)=-1`

Answer: `y^(t)=e^(-3t)`