Saturday

March 28, 2015

March 28, 2015

Posted by **Claire** on Tuesday, February 8, 2011 at 4:40am.

Solve the initial-value problem:

problem:

dy/dx: 4(3cos(6x)-1)/1+2x-sin(6x) y=12 when x=0

- Calc-IVP -
**MathMate**, Tuesday, February 8, 2011 at 9:02amDo you mean

dy/dx= 4(3cos(6x)-1) / (1+2x-sin(6x)) ?

Substitute

u=(1+2x-sin(6x))

du=(-6cos(6x)+2)

so

dy/dx = -2∫du/u

y=-2ln(1+2x-sin(6x))+C

Initial condition: y(0)=12

y(0)=12=-2ln(1+2(0)-sin(6(0)))+C

12=-2ln(1)+C

12=0+C

=> C=12

So

y=-2ln(1+2x-sin(6x))+12

Check the result by differentiation.

**Answer this Question**

**Related Questions**

Maths - Solve the initial value problem (dy/dt)+2y=15 sin(t)+15 cos(t) with y(0...

Math - Directions: Find the Location on Unit Circle, Period & General Solution ...

Calculus - Solve the following initial value problem explicitly: y'= ye^-x , y(0...

Maths - Solve the initial value problem 10(t+1)(dy/dt)+9y= 9t for t > -1 with...

Maths - Solve the initial value problem t(dy/dt)+2y=9t with x(1)=4

Maths - Solve the initial value problem (dx/dt)+2x=cos(2t) with x(0)=5

Calculus - Use a trig identity to combine two functions into one so you can ...

trig - I'm trying to see if the teacher didn't give us the right problem or if I...

maths - solve 3cos^2 2x +4 sin 2x =1 0<x<360

Maths - Solve the initial value problem (dx/dt)-2ty=-12t^2e^t^2 and the ...