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Find the general solution for the differential equation. (dy)/(dx)=8x^(2)y-2xy

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This is a first-order homogeneous differential equation, which has the general form:

dy/dx + P(x)y = 0

In order to solve it, we will first rearrange the equation to match this form:

dy/dx - (2x-8x^2) y = 0

Now, we will apply an integrating factor. The integrating factor is typically expressed as e^(∫P(x)dx).

We calculate the integral of P(x), which is - (2x-8x^2):

∫(2x - 8X^2) dx = x^2 - 8x^3/3 + C

where C is the constant of integration. The integrating factor becomes:

e^(x^2 - 8x^3/3 + C)

We multiply every term in the differential equation by the integrating factor:

e^(x^2 - 8x^3/3 + C) * dy/dx - e^(x^2 - 8x^3/3 + C) *(2x-8x^2)y = 0

Using the property of the derivative of a product of two functions, we express the left hand side as the derivative with respect to x of [e^(x^2 - 8x^3/3 + C) * y]:

d/dx [e^(x^2 - 8x^3/3 + C) * y] = 0

Now, we can solve this differentional equation by direct integration:

∫d/dx [e^(x^2 - 8x^3/3 + C) * y] dx = ∫0 dx

e^(x^2 - 8x^3/3 + C) * y = γ

where γ is the constant of integration.


Finally, expressing y in terms of x, we get the general solution:

y(x) = γ * e^(-x^2 + 8x^3/3 - C)

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User Draco
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