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(B) Divide the given differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$ by $x^2$ to get $\frac{2y^2}{x^2} \frac{dx}{dy} - \frac{2y}{x} + 1 = 0

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$\frac{2}{3} e^2$

Vedclass · Answer

(B) Divide the given differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$ by $x^2$ to get $\frac{2y^2}{x^2} \frac{dx}{dy} - \frac{2y}{x} + 1 = 0$.Let $v = \frac{1}{x}$, then $\frac{dv}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$.Substituting this into the equation, we get $-2y^2 \frac{dv}{dy} - 2yv + 1 = 0$, which simplifies to $\frac{dv}{dy} + \frac{1}{y} v = \frac{1}{2y^2}$.This is a linear differential equation of the form $\frac{dv}{dy} + P(y)v = Q(y)$, where $P(y) = \frac{1}{y}$ and $Q(y) = \frac{1}{2y^2}$.The integrating factor is $IF = e^{\int P(y) dy} = e^{\int \frac{1}{y} dy} = e^{\ln y} = y$.The general solution is $v \cdot IF = \int Q(y) \cdot IF dy + C$, which gives $v \cdot y = \int \frac{1}{2y^2} \cdot y dy + C = \int \frac{1}{2y} dy + C = \frac{1}{2} \ln y + C$.Given $x(e) = e$, we have $v = \frac{1}{e}$ at $y = e$. Substituting these values: $\frac{1}{e} \cdot e = \frac{1}{2} \ln e + C$, so $1 = \frac{1}{2} + C$, which gives $C = \frac{1}{2}$.Thus, $v \cdot y = \frac{1}{2} \ln y + \frac{1}{2} = \frac{\ln y + 1}{2}$.Since $v = \frac{1}{x}$, we have $\frac{y}{x} = \frac{\ln y + 1}{2}$, which implies $x = \frac{2y}{\ln y + 1}$.For $y = e^2$, $x(e^2) = \frac{2e^2}{\ln e^2 + 1} = \frac{2e^2}{2 + 1} = \frac{2}{3} e^2$.

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0, y > 1, x(e) = e$. Then $x(e^2)$ is equal to:

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