Friday

March 27, 2015

March 27, 2015

Posted by **billy** on Saturday, March 19, 2011 at 6:23pm.

- calculus (()) -
**MathMate**, Saturday, March 19, 2011 at 6:53pmWatch 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.

**Answer this Question**

**Related Questions**

calculus - how do you solve the initial value problem by using separation of ...

calculus - how do you solve the initial value problem by using separation of ...

separation of variables - solving the initial value problem by separation of ...

calculus - how do you solve the initial value problem by separation of variables...

calculus - solve the initial value problem by separation of variables dy/dx=-x^...

calculus - solve the initial value problem by separation of variables dy/dx=-xy^...

calculus - solve the initial value problem by separation of variables dy/dx=-xy^...

calculus - solve the initial value problem by separation of variables dy/dx=-x^...

calculus - solve the initial value problem by separation of variables 8. dy/dx=x...

calculus - solve the initial value problem by separation of variables dy/dx=y^(2...