# 1. solve the differential equation dy/dx=x(3+y)

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### 2 Answers

1) You need to separate the variables, hence you need to multiply both sides by `(dx)/(3+y)` such that:

`(dy)/(dx)*(dx)/(3+y)=x(3+y)*(dx)/(3+y)`

You need to reduce like terms such that:

`(dy)/(3+y)=x(dx)`

You need to integrate both sides to find the general solution to differential equation such that:

`int (dy)/(3+y) = int x(dx)`

`ln|3+y| = x^2/2 + c`

`3 + y = e^(x^2/2) + c =gt y = e^(x^2/2+c) - 3 `

**Hence, the general solution to differential equation `(dy)/(dx)=x(3+y)` is `y = e^(x^2/2+c) - 3` .**

2) You need to perform the same first step as the first problem, thus you need to separate the variables, hence you need to multiply both sides by `y(dx)` such that:

`ydy=(tan(5x))dx`

You need to integrate both sides such that:

`int ydy= int (tan(5x))dx`

`y^2/2 = int (sin(5x))/(cos(5x)) dx`

You need to solve`int (sin(5x))/(cos(5x)) dx` , hence, you should come up with the substitution `cos 5x = p` => `-5sin 5x dx = dp`

`sin 5x dx = -(dp)/5`

`int (sin(5x))/(cos(5x)) dx = int (-dp)/(5p) `

`int (-dp)/(5p) = (-1/5)*ln|p|`

`int (sin(5x))/(cos(5x)) dx = (-1/5)*ln|cos 5x| + c`

`y^2/2 = (-1/5)*ln|cos 5x| + c`

`y^2 = (-2/5)*ln|cos 5x| + c`

`y = sqrt((-2/5)*ln|cos 5x| + c)`

**Hence, the general solution to differential equation `(dy)/(dx)=(tan(5x))/y` yields`y = sqrt((-2/5)*ln|cos 5x| + c)` .**

`(dy)/(dx) = x(3+y) `

`==gt (dy)/(3+y) = x dx `

`==gt int (dy)/(3+y) = int x dx `

`==gt ln (y+3) = x^2/2 + C `

`==gt y+3 = e^((x^2/2)+C) `

`==gt y= Ce^(x^2/2) - 3`

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