Initial Value Problems

Itnitial-value problems are solved by applying initial conditions to the general solution of the differential equation. Note that initial conditions are applied only to the general solution and not to the homogeneous solution `y_h`, even though it is `y_h`, that possesses all the arbitrary constants that must be evaluated. The only exception is when the general solution is the homogeneous solution; that is, when the differential equation under consideration is itself homogeneous.

Example 1. Solve `y''+y'-2y=0` , y(0)=1, y'(0)=2.

This is gomogeneous differential equation. Characteristic equation is `r^2+r-2=0` or `(r+2)(r-1)=0` which has roots `r_1=-2,\ r_2=1` .

So, general solution is `y=c_1e^(-2x)+c_2e^x` .

Thus, `y'=-2c_1e^(-2x)+c_2e^x` .

Applying initial consditions gives




Subtracting second equation from first gives `3c_1=-1` or `c_1=-1/3` .

From first equation `c_2=1-c_1=1-(-1/3)=4/3` .

So, solution is `y=-1/3 e^(-2x)+4/3e^x` .

Example 2. Solve `y''+2y'+y=cos(t)` , y(0)=3, y'(0)=0

First solve correspondint homogeneous equation `y''+2y'+y=0` . Characteristic equation is `r^2+2r+1=0` or `(r+1)^2=0` which has roots `r_1=-1, r_2=-1` .

There is one root of multiplicity 2, so solution of homogeneous equation `y_h=c_1e^(-t)+c_2te^(-t)` .

Once again, we don't apply initial conditions on this stage. We must first find general solution (in this case `y_h+y_p` ) and only then apply initial consditions.

To find particular solution use method of undetermined coefficients.

Assume that `y_p=Acos(t)+Bsin(t)` . Then `y_p'=-Asin(t)+Bcos(t)` and `y_p''=-Acos(t)-Bsin(t)` .

Plugging this into equation gives

`-Acos(t)-Bsin(t)+2(-Asin(t)+Bcos(t))+Acos(t)+Bsin(t)=2Bcos(t)-2Asin(t)` .

Equating like terms with `cos(t)` gives


Which has solution `A=0, B=1/2` .

So, `y_p=1/2 sin(t)` .

Thus, general solution is `y=y_h+y_p=c_1e^(-t)+c_2te^(-t)+1/2 sin(t)` .

Now, we can use initial conditions.

For this find derivative of general solution: `y'=-c_1e^(-t)+c_2e^(-t)-c_2te^(-t)+1/2 cos(t)` .


`{(y(0)=3=c_1e^(-0)+c_2*0*e^(-0)+1/2 sin(0)),(y'(0)=0=-c_1e^(-0)+c_2e^(-0)-c_2*0*e^(-0)+1/2 cos(0)):}`



Which has solution `c_1=3,\ c_2=5/2` .

Finally, solution is `y=3e^(-t)+5/2 t e^(-t)+1/2 sin(t)` .