For the primitive integral equation ydx + y^2 | vedclass

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(A) Given the differential equation: $ydx + y^2dy = xdy$.Rearranging the terms to form a linear differential equation in $x$:$ydx = (x - y^2)dy$$\frac

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$3$

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(A) Given the differential equation: $ydx + y^2dy = xdy$.Rearranging the terms to form a linear differential equation in $x$:$ydx = (x - y^2)dy$$\frac{dx}{dy} = \frac{x - y^2}{y} = \frac{x}{y} - y$$\frac{dx}{dy} - \frac{1}{y}x = -y$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -y$.The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.$x \cdot \frac{1}{y} = \int (-y) \cdot \frac{1}{y} dy + C$$\frac{x}{y} = \int -1 dy + C = -y + C$.Using the initial condition $y(1) = 1$,we substitute $x = 1$ and $y = 1$:$\frac{1}{1} = -1 + C \Rightarrow C = 2$.So,the equation is $\frac{x}{y} = -y + 2$,which implies $x = 2y - y^2$.To find $y(-3)$,we set $x = -3$:$-3 = 2y - y^2 \Rightarrow y^2 - 2y - 3 = 0$.$(y - 3)(y + 1) = 0$.Since $y > 0$,we have $y = 3$.

For the primitive integral equation $ydx + y^2dy = xdy$ ; $x \in R$,$y > 0$,$y = y(x)$,$y(1) = 1$,then $y(-3)$ is

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