# `yy' = 4sinx` Find the general solution of the differential equation

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

be evaluated using direct integration. The derivative of y denoted as` y'` can be written as `(dy)/(dx)` then `y'= f(x)` can be expressed as `(dy)/(dx)= f(x)` .

For the problem `yy'=4sin(x)` , we may...

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

be evaluated using direct integration. The derivative of y denoted as` y'` can be written as `(dy)/(dx)` then `y'= f(x)` can be expressed as `(dy)/(dx)= f(x)` .

For the problem `yy'=4sin(x)` , we may apply `y' = (dy)/(dx) ` to set-up the integration:

`y(dy)/(dx)= 4sin(x)` .

or `y dy = 4 sin(x) dx`

Then set-up direct integration on both sides:

`inty dy = int 4 sin(x) dx`

Integration:

Apply Power Rule integration: `int u^n du= u^(n+1)/(n+1) ` on `inty dy` .

Note: `y` is the same as `y^1` .

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

`= y^2/2`

Apply the basic integration property: ` int c*f(x)dx= c int f(x) dx` and basic integration formula for sine function: `int sin(u) du = -cos(u) +C`

`int 4 sin(x) dx= 4int sin(x) dx`

`= -4 cos(x) +C`

Then combining the results for the general solution of differential equation:

`y^2/2 = -4cos(x)+C`

`2* [y^2/2] = 2*[-4cos(x)]+C`

`y^2 =-8cos(x)+C`

`y = +-sqrt(C-8cosx)`

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