Second Order Differential equation

To solve the differential equations of the standard form,

`y''-ay=0`

We use the substitution,

`u=y'`

And so the second order differential equation becomes,

`u'=au`

...which is a first order linear differential equation. This has the standard solution,

`lnu=at+c=at+lnu_0`

Solving this for `u`, we obtain

`u=u_0exp(at)`

and this is inserted into the original differential equation,

`y'=y'_0exp(at)`

Upon a final integration we arrive at,

`y=(y'_0)/(a)exp(at)+c`

We can find the value of c using the particular value `y(0)=y_0`,

`y_0=(y'_0)/(a)+c`

Therefore,

`c=y_0-(y'_0)/(a)`

We can now construct the complete solution,

`y=(y'_0)/(a)exp(at)+y_0-(y'_0)/(a)`