# calculus

posted by .

how do you solve the initial value problem by using separation of variables dy/dx=1+x/(sqrt of y), y(2)=9

• calculus (()) -

Watch missing parentheses, please.

Parentheses are needed to enclose numerators and denominators, otherwise additions and subtractions will take place after the division.

Separate the variables,
dy/dx = (1+x)/sqrt(y)
sqrt(y)dy = (1+x)dx
Integrate:
(2/3)y^(3/2) = x + x²/2 + C
y = [(3/2)(x+x²/2+C)]^(2/3)
from which we can solve for C=14.
so
y(x)=[(3/2)(x+x²/2+14)]^(2/3)

Please check all arithmetic.

## Similar Questions

1. ### calculus

solve the initial value problem by separation of variables dy/dx=y^(2)+1, y(1)=0
2. ### calculus

solve the initial value problem by separation of variables 8. dy/dx=x+1/xy, x>0, y(1)=-4
3. ### calculus

how do you solve the initial value problem by using separation of variables dy/dx=1/y^2, y(0)=4
4. ### calculus

solve the initial value problem by separation of variables dy/dx=-x^2y^2, y(4)=4
5. ### calculus

solve the initial value problem by separation of variables dy/dx=-xy^2, y(1)=-0.25
6. ### separation of variables

solving the initial value problem by separation of variables dy/dx=(4(sqrt of y)lnx))/x, y(e)=9
7. ### calculus

solve the initial value problem by separation of variables dy/dx=-xy^2, y(1)=-0.25
8. ### calculus

solve the initial value problem by separation of variables dy/dx=-x^2y^2, y(4)=4
9. ### calculus

how do you solve the initial value problem by separation of variables dy/dx=-x^2y^2, y(4)=4
10. ### calculus

how do you solve the initial value problem by using separation of variables dy/dx=-xy^2 when y(1)=-0.25

More Similar Questions