Let y=f(x) be a particular solution to the differential equation dy/dx=xy^3 with f(1)=2. Write an equation for the tangent line to the graph of y=f(x) at x=1.

Question

darkigloo · Answer

should this be my first step? 
int (y^-3)dy = int x dx

loser66 · Answer

I think so

darkigloo · Answer

and then would i have to find c then plug in c to find y?

loser66 · Answer

after take integral both sides , what do you have?

darkigloo · Answer

$$ \frac{ -1}{ 2y^2 }=\frac{ x^2 }{ 2} + C$$

loser66 · Answer

solve for y,

darkigloo · Answer

$$(\frac{ -2 }{ x^2}+ \frac{ 1 }{ C } ) / 2 = y ^{2}$$

loser66 · Answer

y =$\pm \sqrt{thewholething}$ replace x =1 y =2 to solve for C

darkigloo · Answer

c=1/6?

loser66 · Answer

actually, from the beginning, you have y =f(1) =2 mean point (1,2) is on the graph and you have to find out the tangent line at this point, right?
One more thing, from the original problem , they give you dy /dx = xy^3 which is the slope of the tangent line of the curve defined by xy^3, so, at x =1, y=2, dy/dx =m = 8
you have the slope, and the point, just write out the tangent line by y = mx + b

loser66 · Answer

solving ODE in the case they ask another question for equation of the curve. If they just ask the tangent line, do without solving the ODE , you still have the result.

darkigloo · Answer

ohh ok. so the answer is f(x)=2+8(x-1)

loser66 · Answer

yup, should make it clear by $$\Large y_{tangent line} = 8x -6$$

darkigloo · Answer

ok thank you!