# solve the initial value problem dy/dx = sin(4x)/(3+cos(4x)) for y = 5 when x = 0I have relisted this as I forgot the brackets around the denominator of the fraction, which I am sure will have a...

solve the initial value problem dy/dx = sin(4x)/(3+cos(4x)) for y = 5 when x = 0

I have relisted this as I forgot the brackets around the denominator of the fraction, which I am sure will have a bearing on the reasult.

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### 1 Answer

You need to multiply by dx both sides such that:

`dy = (sin(4x)dx)/(3+cos(4x))`

You need to integrate both sides such that:

`int dy = int (sin(4x)dx)/(3+cos(4x))`

`y = int (sin(4x)dx)/(3+cos(4x))`

You should solve the integral by substitution method need , hence you need to substitute `t ` for `3+cos(4x).`

`t = 3+cos(4x)`

Differentiating both sides yields:

`dt = -4sin(4x)dx =gt sin(4x)dx = -(dt)/4`

You need to write the function `(sin(4x)dx)/(3+cos(4x))` in terms of t such that:

`-(dt)/(4t)` = `(sin(4x)dx)/(3+cos(4x))`

Hence, integrating yields:

`int (sin(4x)dx)/(3+cos(4x)) = int -(dt)/(4t) = -(1/4)*ln|t| + c`

You need to substitute `3+cos(4x)` for `t` in `-(1/4)*ln|t| + c` such that:

`y = -(1/4)*ln|3+cos(4x)| + c`

You need to substitute 0 for x and 5 for y such that:

`5 = -(1/4)*ln|3+cos(0)| + c =gt 5 = -(1/4)*ln|3+1| + c`

`5 = -(1/4)*ln|4| + c =gt c = 5 + ln (4^(1/4))`

`c = 5 + ln root(4) 4`

**Hence, the particular solution to the initial value problem is `y = -(1/4)*ln|3+cos(4x)| + 5 + ln root(4) 4` **