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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​

1 Answer

4 votes

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

User Andrew Larsen
by
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