Question

If y(x) is the solution of the differential equation


`\rm{xdy - ( {y}^(2) - 4y)dy = 0 \: for \: x > 0, \: \: \: \: \: \: \: y(1) = 2,}`
& the slope of the curve y=y(x) is never zero, then the value of
`\rm{10y( √(2) )}` is​

Answer 1

I assume the equation is

`x \, dy - (y^2 - 4y) \, dx = 0`

since separating variables leads to

`x\,dy = y(y-4) \, dx`

`(dy)/(y(y-4)) = \frac{dx}x`

for which the condition that
`x>0` is actually relevant, as opposed to the simpler differential equation

`x \, dx - (y^2-4y)\, dy = 0 \implies y(y-4) \, dy = x \, dx`

(though it's a bit more work to solve for
`y(x)` in this case)

That the slope
`(dy)/(dx)` is non-zero tells us that

`(dy)/(dx) = \frac{y(y-4)}x \\eq 0 \implies y\\eq0 \text{ and } y \\eq 4`

Integrate both sides.

`\displaystyle \int (dy)/(y(y-4)) = \int \frac{dx}x`

On the left, expand into partial fractions.

`\displaystyle \frac14 \int \left(\frac1{y-4} - \frac1y\right) \, dy = \int \frac{dx}x`

`\frac14 (\ln|y-4| - \ln|y|) = \ln|x| + C`

With the given initial value, we find

`y(1) = 2 \implies \frac14 (\ln|2-4| - \ln|2|) = \ln|1| + C \implies C = 0`

so the particular solution is

`\frac14 (\ln|y-4| - \ln|y|) = \ln|x|`

By definition of absolute value, with the initial condition of
`0 < y=2 < 4` and the condition
`x>0`, we can remove the absolute values.

`\frac14 (\ln(4-y) - \ln(y)) = \ln(x)`

Solve for
`y`.

`\ln\left(\frac{4-y}y\right) = 4 \ln(x) = \ln\left(x^4\right)`

`\frac{4-y}y = \frac4y - 1 = x^4`

`\implies y(x) = \frac4{1 + x^4}`

Then

`10y\left(\sqrt2\right) = (40)/(1 + \left(\sqrt2\right)^4) = \boxed{8}`

On the off-chance you meant the other equation I suggested, we find

`\displaystyle \int y(y-4) \, dy = \int x \, dx`

`\displaystyle \frac{y^3}3 - 2y^2 = \frac{x^2}2 + C`

`y(1) = 2 \implies \frac83 - 2\cdot4 = \frac12 + C \implies C = -\frac{35}6`

Solving for
`y(x)` involves picking the right branch of the cube root that agrees with
`y(1)=2`. With the cube root formula, we find

`y(x) = 2 - \xi(1 - i\sqrt3) - \frac1\xi (1+i\sqrt3)`

where

`\xi = \frac{2\sqrt[3]{4}}{\sqrt[3]{3x^2 - 3 + √(9x^4 - 18x^2 - 1015)}}`

With a calculator, we find

`10y\left(\sqrt2\right) \approx 18.748`