Calculus help? Please?

Find the equation for y, given the derivative and the indicated point on the curve. dy/dx = 3x^2-5; y(0)=1

Question

Calculus help? Please?

Find the equation for y, given the derivative and the indicated point on the curve. dy/dx = 3x^2-5; y(0)=1

ganeshie8 · Answer

take integral both sides

ganeshie8 · Answer

dy/dx = 3x^2-5 

y = integral  (3x^2-5) dx

anonymous · Answer

How do I integrate that one? do I use x^n dx = x^(n+1)/(n+1) +C?

anonymous · Answer

@ganeshie8

anonymous · Answer

@mathmale ?

mathmale · Answer

Hello, VBG!!
I view this as a separable differential equation.  You comfortable with that?  If so, please separate x-terms from y-terms and write out the result.

mathmale · Answer

BRB

anonymous · Answer

If thats what you think I agree. I learn from you guys! :) Ok so I should solve 3x^2 and -5 separately?

anonymous · Answer

integrate I mean

mathmale · Answer

Let me type that out in Equation Editor for added clarity:

anonymous · Answer

ok

mathmale · Answer

$$\frac{ dy }{ dx}=3x ^{?}-5$$

mathmale · Answer

becomes $$dy=(3x ^{2}-5)dx$$

mathmale · Answer

Why is this advantageous?

anonymous · Answer

advantageous? because it is easier to solve?

mathmale · Answer

Yes, VBG, you do need to integrate 3x^2 and -5 separately.  Add a constant of integration as well.

mathmale · Answer

I'd prefer you explain that what I did is advantageous because it emphasizes that yours is a separable differential equation.

mathmale · Answer

$$\int\limits_{}^{}dy = \int\limits_{}^{}(3x ^{2}-5)dx$$

anonymous · Answer

right ok. so integral 3x^2 = x^3 + C

anonymous · Answer

and integral -5 = -5x + C

anonymous · Answer

is this correct?

anonymous · Answer

mathamle I have to go and will be back in an hour or so!! hopefully you will be on :)

anonymous · Answer

@mathmale Hi sorry for taking so long!!

mathmale · Answer

The right side of your integral now becomes x^3-5x + C.  What's the left side?

The initial condition says y(0)=1.  In y=x^3-5x + c, let x=0 and y=1.  Find C.
Then write the solution in the form y=x^3-5x + C.

anonymous · Answer

the left side would just show the integral x^n dx?

Ok so y= 0^3 - 5(0) + C
So C= 1

anonymous · Answer

yes?

mathmale · Answer

Then y=x^3-5x + 1.
Check this by substitution of y=x^3 - 5x + 1 and y ' = 3x^2 -5 back into the original equation.
Sorry for the delay; I messed up a display setting on my computer and had to re-learn how to correct that.  Hope to see you again soon on OpenStudy!

anonymous · Answer

yep thats what I did.
No problem thanks!