Slope of a curve and at a point

[yecko]

Homework Statement

http://i.imgur.com/In40pGm.png

Answer: C

Homework Equations

f'(x)=slope=(y1-y2)/(x1-x2)

The Attempt at a Solution

I can't even list a valid formula for that...
like I tried to integrate f'(x), but f(x) is with y so I don't think I am thinking in the right direction.
What are the steps in order to get the correct answer?
Thank you very much.

 

[Ssnow]

It is a plane curve? If it is $y=f(x)$ so you can write $\frac{f'(x)}{f(x)^2}=\frac{1}{x^3}$ and integrating both members you will find the equation of $f(x)$ ...
Ssnow

 

[Ray Vickson]

Homework Statement

http://i.imgur.com/In40pGm.png
View attachment 200123
Answer: C

Homework Equations

f'(x)=slope=(y1-y2)/(x1-x2)

The Attempt at a Solution

I can't even list a valid formula for that...
like I tried to integrate f'(x), but f(x) is with y so I don't think I am thinking in the right direction.
What are the steps in order to get the correct answer?
Thank you very much.

Solve the differential equation
$$\frac{dy}{dx} = \frac{y^2}{x^3}$$
The general solution $y(x)$ will contain an unknown constant $c$, whose value can be obtained by using the given condition $y(1) = 1$.

 

[yecko]

I know that (1,1) is for solving the constant. however, i can't find the equation.
f'(x)=y^2/x^3
y=f(x)=y^2/x^2*(-1)+c
This equation seems unreasonable...
can you help me in this?
thanks

 

[Ssnow]

You must integrate the differential equation with the separable variables method ..

 

[yecko]

Like what integration method? I can't think of any seems applicable..
Any suggestion? Thanks

 

[vela]

You have to rewrite the differential equation so all the x's are on one side and all the y's are on the other. Then you can integrate each side.

 

[Ssnow]

Solve the two integrals $\int \frac{dy}{y^2} =\int \frac{dx}{x^{3}}+c$, you will obtain something as $y(x)= ...(x)+c$...
Ssnow

 

[yecko]

∫dyy2=∫dxx3+c

how does this come?

You have to rewrite the differential equation so all the x's are on one side and all the y's are on the other. Then you can integrate each side.

y'=f'(x)=y^2/x^3
y=f(x)=y^2/x^2*(-2)+c
1/y=1/x^2*(-2)+c
you mean like this? or have I calculated anything wrong?

 

[Ssnow]

From $\frac{dy}{dx}=\frac{y^2}{x^3}$ you treat formally $dx$ and $dy$ as certain ''quantities''. Algebraically you can separate $x$ from $y$ obtaining

$\frac{dy}{y^2}=\frac{dx}{x^3}$
after you can integrate both members ...
see https://en.wikipedia.org/wiki/Separation_of_variables
Ssnow

 

[yecko]

http://i.imgur.com/cru3f8g.jpg

Still I don't think I have done the right thing...
any more help? Thanks

 

[mjc123]

You can't go from 1/y = 1/2x² to y = 2x² + C. It must be 1/y = 1/2x² + C. You add the integration constant in the integration step, not after any subsequent manipulations. Then y = 2x²/(1+2Cx²).

 

[yecko]

thank you for all of your help
i can finally solved it

 

[Ssnow]

Yes, as @mjc123 said you must put the constant $C$ after the second integration and consider it in all algebraic passages...

Ssnow

 

[Niladri Dan]

Is it a problem of RSM?