How can I integrate x^3/y^2 to solve this differential equation?

AI Thread Summary
To solve the differential equation dy/dx = x^3/y^2 with the initial condition y(2) = 3, the equation can be separated into x^3dx = y^2dy. After separating, both sides can be integrated, resulting in an equation that includes a constant of integration. The initial condition y(2) = 3 will help determine the value of this constant. This approach simplifies the integration process and leads to the solution of the differential equation.
Pseudo Statistic
Messages
391
Reaction score
6
How can I find the solution to this differential equation: dy/dx = x^3/y^2 given y(2) = 3?
I'd just like a hint on how I can integrate x^3/y^2... because that's where I'm falling.
Thanks for any responses.
 
Physics news on Phys.org
Can you not separate the equation and then integrate?
 
When I seperate, do I get:
x^3dx - y^2dy = 0?
Or is this wrong?
I was asking to check..
Thanks.
 
Pseudo Statistic said:
When I seperate, do I get:
x^3dx - y^2dy = 0?
Or is this wrong?
I was asking to check..
Thanks.

That's fine- although you might find it simpler yet to write the equation as
x3dx= y2dy

Now integrate both sides. You get a "constant of integration" when you do that. Remember that y(2)= 3 means that when x= 2, y= 3. That will help you determine what that constant must be.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top