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

enter image source here

2 Answers

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.

May 14, 2016

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