Slope of a curve and at a point

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SUMMARY

The discussion centers on solving the differential equation \(\frac{dy}{dx} = \frac{y^2}{x^3}\) to find the function \(y(x)\). Participants emphasize the importance of separating variables before integrating both sides, leading to the general solution that includes an unknown constant \(c\). The correct approach involves integrating \(\int \frac{dy}{y^2} = \int \frac{dx}{x^3} + c\) and properly applying the integration constant during the solution process. Ultimately, the solution is confirmed as \(y = \frac{2x^2}{1 + 2Cx^2}\).

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  • Understanding of differential equations, specifically separable variables
  • Familiarity with integration techniques, particularly for rational functions
  • Knowledge of initial conditions and their role in determining constants in solutions
  • Basic calculus concepts, including derivatives and slopes of curves
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  • Learn advanced integration techniques for rational functions
  • Explore initial value problems and their applications in differential equations
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yecko
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Homework Statement


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

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##.
 
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
 
You must integrate the differential equation with the separable variables method ..
 
Like what integration method? I can't think of any seems applicable..
Any suggestion? Thanks
 
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.
 
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
 
Ssnow said:
∫dyy2=∫dxx3+c
how does this come?
vela said:
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?
 
  • #10
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
 
  • #11
  • #12
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).
 
  • Like
Likes Ssnow
  • #13
thank you for all of your help
i can finally solved it
 
  • #14
Yes, as @mjc123 said you must put the constant ##C## after the second integration and consider it in all algebraic passages...

Ssnow
 
  • #15
Is it a problem of RSM?
 

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