Show that substituting y = xv, where v is a function of x, in the differential equation "xy(dy/dx) + y^2 − 2x^2 = 0" (with x is not equal to 0) leads to the differential equation "xv(dv/dx) + 2v^2 − 2 = 0"

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This is the first part of a Step 1 question (2012 question 8), and is fairly typical in that it requires A Level Maths understanding to be applied in a number of different ways, out of the usual context.

We begin by finding a new expression for (dy/dx), in terms of v. We do this by differentiating "y = xv", and getting:

dy/dx = x(dv/dx) + v

By substituting this and "y = xv" into the original differential equation, we get:

(x)(xv)(x(dv/dx) + v) + (xv)^2 - 2(x^2) = 0

(x^3)v(dv/dx) + 2(xv)^2 - 2(x^2) = 0

We can divide by x^2, since we know that x does not equal zero. This gives the required result:

xv(dv/dx) + 2(v^2) - 2 = 0

The rest of the question first asks you to find a solution to the original equation. The new equation is now separable, and can be solved to find v. This can be substituted back into "y = xv", giving a solution for y. Try it if you like. It should look something like:

(x^2)(y^2 - x^2) = C (where C is a constant)

The question then asks you to solve (again with x is not equal to zero):

y(dy/dx) + 6x + 5y = 0

This can be done with the same substitution, which again makes it separable. Using partial fractions, integrating, finding v in terms of x (or vice versa), and substituting allows a solution to be found. This should (although it is still pretty ugly, particularly on a computer screen) look like:

((y + 3x)^3)/((y + 2x)^2) = B (where B is a constant)

Will W. GCSE Maths tutor, A Level Maths tutor, A Level Further Mathem...

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