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saul85 [17]
4 years ago
11

The differential equation xy' = y(in x – Iny) is neither separable, nor linear. By making the substitution y(x) = xv(x), show th

at the new equation for v(x) equation is separable. N.B. you do not have to actually solve the ODE.
Mathematics
1 answer:
FinnZ [79.3K]4 years ago
8 0

Answer:

We can place everything with v(x) on one side of the equality, everything with x on the other side. This is done on the step-by-step explanation, and shows that the new equation is separable.

Step-by-step explanation:

We have the following differential equation:

xy' = y(ln x - ln y)

We are going to apply the following substitution:

y = xv(x)

The derivative of y is the derivative of a product of two functions, so

y' = (x)'v(x) + x(v(x))'

y' = v(x) + xv'(x)

Replacing in the differential equation, we have

xy' = y(ln x - ln y)

x(v(x) + xv'(x)) = xv(x)(ln x - ln xv(x))

Simplifying by x:

v(x) + xv'(x) = v(x)(ln x - ln xv(x))

xv'(x) = v(x)(ln x - ln xv(x)) - v(x)

xv'(x) = v(x)((ln x - ln xv(x) - 1)

Here, we have to apply the following ln property:

ln a - ln b = ln \frac{a}{b}

So

xv'(x) = v(x)((ln \frac{x}{xv(x)} - 1)

Simplifying by x,we have:

xv'(x) = v(x)((ln \frac{1}{v(x)} - 1)

Now, we can apply the above ln property in the other way:

xv'(x) = v(x)(ln 1 - ln v(x) -1)

But ln 1 = 0

So:

xv'(x) = v(x)(- ln v(x) -1)

We can place everything that has v on one side of the equality, everything that has x on the other side, so:

\frac{v'(x)}{v(x)(- ln v(x) -1)} = \frac{1}{x}

This means that the equation is separable.

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