If y(x) is the solution of the differential e | vedclass

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(B) Given the differential equation: $x dy = (y^2 - 4y) dx$ for $x > 0$. Separating the variables,we get: $\int \frac{dy}{y^2 - 4y} = \int \frac{dx}{x

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Given the differential equation: $x dy = (y^2 - 4y) dx$ for $x > 0$. Separating the variables,we get: $\int \frac{dy}{y^2 - 4y} = \int \frac{dx}{x}$. Using partial fractions: $\int \frac{1}{4} (\frac{1}{y-4} - \frac{1}{y}) dy = \int \frac{dx}{x}$. Multiplying by $4$: $\int (\frac{1}{y-4} - \frac{1}{y}) dy = 4 \int \frac{dx}{x}$. Integrating both sides: $\ln|y-4| - \ln|y| = 4 \ln x + \ln c$. This simplifies to: $\ln|\frac{y-4}{y}| = \ln(cx^4)$,which implies $\frac{y-4}{y} = cx^4$. Given $y(1) = 2$,we substitute $x=1, y=2$: $\frac{2-4}{2} = c(1)^4 \Rightarrow c = -1$. So,$\frac{y-4}{y} = -x^4 \Rightarrow y-4 = -yx^4 \Rightarrow y(1+x^4) = 4 \Rightarrow y = \frac{4}{1+x^4}$. We need to find $10y(\sqrt{2})$. $y(\sqrt{2}) = \frac{4}{1+(\sqrt{2})^4} = \frac{4}{1+4} = \frac{4}{5}$. Therefore,$10y(\sqrt{2}) = 10 	imes \frac{4}{5} = 8$.

If $y(x)$ is the solution of the differential equation $x dy - (y^2 - 4y) dx = 0$ for $x > 0$ and $y(1) = 2$,and the slope of the curve $y = y(x)$ is never zero,then the value of $10y(\sqrt{2})$ is . . . .

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