Show that # y^2 = (4x)(a-x) = 4ax-4x^2 # is a solution to the DE? # 2xy dy/dx = y^2 - 4x^2#
(portions of this question have been edited or deleted!)
(portions of this question have been edited or deleted!)
As the quoted Differential Equation is malformed, let us work back from the quoted solution to form the correct DE:
Correction of the Question:
If a solution is:
is a solution, then implicit differentiation gives:
Therefore the question should (presumably) be to solve the DE
Solution:
We have:
As suggested, let us perform the substitution:
Substituting into the (corrected) DE [A] we have:
This is now a First Order separable DE, so we can "separate the variables" to get:
And now we can integrate, to get:
Applying the initial conditions:
So our particular solution is:
Restoring the substitution:
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To show that (y^2 = (4x)(a-x) = 4ax - 4x^2) is a solution to the differential equation (2xy \frac{dy}{dx} = y^2 - 4x^2):
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Calculate ( \frac{dy}{dx} ) and ( \frac{d^2y}{dx^2} ). [ \frac{dy}{dx} = \frac{d}{dx}(4ax - 4x^2) = 4a - 8x ] [ \frac{d^2y}{dx^2} = \frac{d}{dx}(4a - 8x) = -8 ]
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Plug (y^2 = 4ax - 4x^2) and its derivatives into the given differential equation. [ 2xy \frac{dy}{dx} = 2x(4ax - 4x^2)(4a - 8x) = 2x(16ax - 8ax^2 - 32x + 8x^2) = 32ax^2 - 16ax^3 - 64x^2 + 16x^3 ] [ = 16x^3 - 16ax^3 - 64x^2 + 32ax^2 = -16ax^3 + 32ax^2 - 64x^2 + 16x^3 ]
[ y^2 - 4x^2 = (4ax - 4x^2)^2 - 4x^2 = (16a^2x^2 - 32ax^3 + 16x^4) - 4x^2 ] [ = 16a^2x^2 - 32ax^3 + 16x^4 - 4x^2 ] [ = 16x^4 - 32ax^3 + 16a^2x^2 - 4x^2 ]
- Check if (2xy \frac{dy}{dx} = y^2 - 4x^2) is true for (y^2 = 4ax - 4x^2). [ -16ax^3 + 32ax^2 - 64x^2 + 16x^3 = 16x^4 - 32ax^3 + 16a^2x^2 - 4x^2 ]
The equation holds true. Therefore, (y^2 = 4ax - 4x^2) is a solution to the given differential equation.
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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.
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.
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