Let alpha x = exp(x^beta y^gamma) be the solu | vedclass

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(A) Given the differential equation: $2x^2 y \frac{dy}{dx} - 1 + xy^2 = 0$.Let $y^2 = t$. Then $2y \frac{dy}{dx} = \frac{dt}{dx}$.Substituting this in

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(A) Given the differential equation: $2x^2 y \frac{dy}{dx} - 1 + xy^2 = 0$.Let $y^2 = t$. Then $2y \frac{dy}{dx} = \frac{dt}{dx}$.Substituting this into the equation,we get: $x^2 \frac{dt}{dx} + xt = 1$.Dividing by $x^2$,we get the linear differential equation: $\frac{dt}{dx} + \frac{1}{x} t = \frac{1}{x^2}$.The integrating factor is $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.The solution is $t \cdot x = \int \frac{1}{x^2} \cdot x dx + C = \int \frac{1}{x} dx + C = \ln x + C$.Substituting $t = y^2$,we have $xy^2 = \ln x + C$.Given $y(2) = \sqrt{\ln 2}$,so $y^2(2) = \ln 2$. Substituting $x=2$ and $y^2=\ln 2$: $2(\ln 2) = \ln 2 + C$,which gives $C = \ln 2$.Thus,$xy^2 = \ln x + \ln 2 = \ln(2x)$.Taking the exponential of both sides: $e^{xy^2} = 2x$,or $2x = \exp(x^1 y^2)$.Comparing this with $\alpha x = \exp(x^\beta y^\gamma)$,we get $\alpha = 2$,$\beta = 1$,and $\gamma = 2$.Therefore,$\alpha + \beta - \gamma = 2 + 1 - 2 = 1$.

Let $\alpha x = \exp(x^\beta y^\gamma)$ be the solution of the differential equation $2x^2 y \frac{dy}{dx} - (1 - xy^2) = 0$,for $x > 0$ and $y(2) = \sqrt{\log_e 2}$. Then $\alpha + \beta - \gamma$ equals:

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