Question

How to solve the separable differential equation and find the particular solution satisfying the initial condition y(−4)=3 ?

Answer 1

General Solution: `color(red)((4y+13)^(1/2)-2x=C_1)" "`
Particular Solution: `color(blue)((4y+13)^(1/2)-2x=13)`

Explanation:

From the given differential equation `y'(x)=sqrt(4y(x)+13)`

take note, that `y' (x)=dy/dx` and `y(x)=y`, therefore

`dy/dx=sqrt(4y+13)`

divide both sides by `sqrt(4y+13)`

`dy/dx(1/sqrt(4y+13))=sqrt(4y+13)/sqrt(4y+13)`

`dy/dx(1/sqrt(4y+13))=1`

Multiply both sides by `dx`

`dx*dy/dx(1/sqrt(4y+13))=dx*1`

`cancel(dx)*dy/cancel(dx)(1/sqrt(4y+13))=dx*1`

`dy/sqrt(4y+13)=dx`

transpose `dx` to the left side

`dy/sqrt(4y+13)-dx=0`

integrating on both sides we have the following results

`int dy/sqrt(4y+13)-int dx=int 0`

`1/4*int (4y+13)^(-1/2)*4*dy-int dx=int 0`

`1/4*(4y+13)^(-1/2+1)/((1-1/2))-x=C_0`

`1/2*(4y+13)^(1/2)-x=C_0`

`(4y+13)^(1/2)-2x=2*C_0`

`color(red)((4y+13)^(1/2)-2x=C_1)" "`General Solution

But `y(-4)=3` means when `x=-4` , `y=3`
We can now solve for `C_1` to solve for the particular solution

`(4y+13)^(1/2)-2x=C_1`

`(4(3)+13)^(1/2)-2(-4)=C_1`

`C_1=13`

Therefore , our particular solution is

`color(blue)((4y+13)^(1/2)-2x=13)`

God bless....I hope the explanation is useful.

Answer 2

`y=x^2+13x+36`, with `y>=-13/4`.

Explanation:

`y>=-13/4`, to make `sqrt(4y+13)` real..

Rearranging,

`x'(y)=1/sqrt(4y+13)`

So, `x=int 1/sqrt(4y+13) dy`

`=(4/2)sqrt(4y+13) + C`

Using `y =3, when x = -4, C = -`13/2`

So. `x = (1/2)(sqrt(4y+13) - 13)`

Inversely. `y = (1/4)((2x+13)^2 - 13) = x^2+13x + 36`