If y=y(x) is the solution curve of the differ | vedclass

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(A) Given the differential equation: $(x^2-4) dy = (y^2-3y) dx$.Separating the variables,we get: $\int \frac{dy}{y(y-3)} = \int \frac{dx}{x^2-4}$.Usin

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$\frac{3}{1+(8)^{1/4}}$

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(A) Given the differential equation: $(x^2-4) dy = (y^2-3y) dx$.Separating the variables,we get: $\int \frac{dy}{y(y-3)} = \int \frac{dx}{x^2-4}$.Using partial fractions: $\frac{1}{3} \int (\frac{1}{y-3} - \frac{1}{y}) dy = \frac{1}{4} \ln |\frac{x-2}{x+2}| + C$.Integrating: $\frac{1}{3} \ln |\frac{y-3}{y}| = \frac{1}{4} \ln |\frac{x-2}{x+2}| + C$.Given $y(4) = \frac{3}{2}$,substitute $x=4$ and $y=\frac{3}{2}$:$\frac{1}{3} \ln |\frac{3/2-3}{3/2}| = \frac{1}{4} \ln |\frac{4-2}{4+2}| + C \Rightarrow \frac{1}{3} \ln |-1| = \frac{1}{4} \ln |\frac{1}{3}| + C \Rightarrow 0 = -\frac{1}{4} \ln 3 + C \Rightarrow C = \frac{1}{4} \ln 3$.Now,for $x=10$: $\frac{1}{3} \ln |\frac{y-3}{y}| = \frac{1}{4} \ln |\frac{10-2}{10+2}| + \frac{1}{4} \ln 3 = \frac{1}{4} \ln |\frac{8}{12}| + \frac{1}{4} \ln 3 = \frac{1}{4} \ln |\frac{2}{3} 	imes 3| = \frac{1}{4} \ln 2$.So,$\ln |\frac{y-3}{y}| = \frac{3}{4} \ln 2 = \ln (2^{3/4}) = \ln (8^{1/4})$.Since $y(4) = 1.5$ and the slope is never zero,$y$ remains in $(0, 3)$,so $\frac{y-3}{y} = -8^{1/4}$.$y-3 = -y \cdot 8^{1/4} \Rightarrow y(1+8^{1/4}) = 3 \Rightarrow y = \frac{3}{1+8^{1/4}}$.

If $y=y(x)$ is the solution curve of the differential equation $(x^2-4) dy-(y^2-3y) dx=0$,$x>2$,$y(4)=\frac{3}{2}$ and the slope of the curve is never zero,then the value of $y(10)$ equals:

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