# How do you solve the differential equation #dy/dx=6y^2x#, where #y(1)=1/25# ?

We are given:

Separate the variables:

Integrate both sides:

Hence, the final solution is:

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The solution to the differential equation ( \frac{dy}{dx} = 6y^2x ), with the initial condition ( y(1) = \frac{1}{25} ), is ( y(x) = \frac{1}{1+3x^3} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- If #y=ln(2x^2-6x)#, then what is #dy/dx#?
- What is a general solution to the differential equation #dy/dx=10-2y#?
- How do you find the exact area of the surface obtained rotating the curve about the #x#-axis of #y=sqrt(8-x)#, #2<=x<=8#?
- How do you find the length of the curve #y=sqrtx-1/3xsqrtx# from x=0 to x=1?
- How do you find the volume of the solid obtained by rotating the region bounded by the curves #y=x#, #x=0#, and #y=(x^2)-6# rotated around the #y=3#?

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