# Thread: Initial value problem

1. ## Initial value problem

hi

got the problem solve the initial value problem

dy/dx=(sin(5x)+6x)/(4-cos(5x)+15x^2) when y=7 and x=0

thought i would try and use formula integration f'(x)/f(x) dx= ln(f(x))+c

i got (-1/5cos(5x)+3x^2)/(4-cos(5x)+15x^2)

= -5ln(4-cos(5x)+15x^2)

2. $\displaystyle \frac{dy}{dx}=\frac{(sin(5x)+6x)}{(4-cos(5x)+15x^2)}$

$\displaystyle \frac{dy}{dx}=\frac{1}{5} \frac{5(sin(5x)+6x)}{(4-cos(5x)+15x^2)}$

$\displaystyle y=\frac{1}{5}\int \frac{5(sin(5x)+6x)}{(4-cos(5x)+15x^2)}$

Using the formula for integration: $\displaystyle \int \frac{f'(x)}{f(x)} dx= ln(f(x))+C$, we get

$\displaystyle y=\frac{1}{5} ln (4-cos(5x)+15x^2)+C$

At $\displaystyle x=0,\ y=7$

$\displaystyle 7=\frac{1}{5} ln (4-1)+C$

$\displaystyle 7=\frac{1}{5} ln 3+C$

$\displaystyle C=7-\frac{1}{5}ln 3$

$\displaystyle y=\frac{1}{5} ln (4-cos(5x)+15x^2)+7-\frac{1}{5}ln 3$

$\displaystyle y=\frac{1}{5} ln (\frac{4-cos (5x)+15x^2}{3})+7$

3. many thanks for the clear presentation of your working.

thanks

4. ## But... where does the rest of the equation "go"?

Sorry, this is probably a silly question, but where does the 1/5 & the 5 come from (second image - 1/5 ((5 sin (5x)+6x)/(4-cos(5x)+15x^2))?

And why does (5 sin(5x)+6x) then disappear? Is something cancelling it out, or am I just not understanding the formula completely?

It's so confusing!!

Many thanks for your help,

S

5. You need the 1/5 because the derivative of 4-cos(5x)+15x^2 is 5sin(5x)+30x which is 5 times more than the what you are given so you multiply what you are given by 5 and multiply the whole equation by 1/5. Then using the constant multiple rule you can take the 1/5 outside when you integrate.