Question

Given that y=2 when x=-3,What is the value of y when x=5 for the differential equation y dy/dx =2x+3?

Answer 1

`y^2 = 2x^2 +6x + 4`

`x=5 => y = +- 2sqrt(21)`

Explanation:

We have:

`ydy/dx=2x+3` with `y(-3)=2`

The DE is separable, so we simply separate the variables" to get:

`int \ y \ dy = int \ 2x+3 \ dx`

And Integrating we get:

`1/2y^2 = x^2 +3x + C`

Using the initial condition, `y(-3)=2`, we have:

`1/2(2^2) = (-3)^2 +3(-3) + C`
`\ \ \ \ \ => 2 = 9 - 9 + C`
`\ \ \ \ \ => C=2`

Thus, we gain the particular solution:

`1/2y^2 = x^2 +3x + 2`
`\ \ \ \ \ => y^2 = 2x^2 +6x + 4`

And when `x=5` we have:

`y^2 = 2(5^2) +6(5) + 4`
`\ \ \ \ =50 +30 + 4`
`\ \ \ \ = 84`

And so there are two solutions:

`y=+-sqrt(84) = +- 2sqrt(21)`