`yy' = -8cospix` Find the general solution of the differential equation

Expert Answers
marizi eNotes educator| Certified Educator

The general solution of a differential equation in a form of` f(y) y'=f(x)` can

 be evaluated using direct integration.

We can denote `y'` as `(dy)/(dx) ` then, 

`f(y) y'=f(x)`

`f(y) (dy)/(dx)=f(x)`

Rearrange into : `f(y) (dy)=f(x) dx`

To be able to apply direct integration : `intf(y) (dy)=int f(x) dx.`

 Applying this to the given problem: `yy'=-8cos(pix)` ,  we get:



`int y(dy)=int-8cos(pix)dx`

For the integration on the left side, we apply Power Rule integration: int u^n `du= u^(n+1)/(n+1)` on int `y dy` .

`int y dy = y^(1+1)/(1+1)`

            `= y^2/2`

For the integration on the right side, we apply the basic integration property: `int c*f(x)dx= c int f(x) dx` and basic integration formula for cosine function: `int cos(u) du = sin(u) +C`

`int -8 cos(pix) dx= -8 int cos(pix) dx`

Let `u = pix` then `du = pi dx` or` (du)/pi=dx.`

Then the integral becomes:

`-8 int cos(pix) dx=-8 int cos(u) *(du)/pi`

                              `=-8/pi int cos(u) du`

                             `=-8/pi*sin(u) +C`

Plug-in `u=pix` in `-8/pi*sin(u) +C` , we get:

`-8 int cos(pix) dx=-8/pi*sin(pix) +C`


Combing the results, we get the general solution for differential equation `(yy'=-8cos(pix))` as:

`y^2/2=-8/pi*sin(pix) +C`

`2* [y^2/2] = 2*[-8/pi*sin(pix)]+C`

`y^2 =-16/pi*sin(pix)+C`

The general solution:` y ^2=-16/pisin(pix)+C` can be expressed as:

`y = +-sqrt(-16/pisin(pix)+C)` .