A solution curve of the differential equation | vedclass

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(D) The given differential equation is $(x+2)(x+2+y) \frac{dy}{dx}-y^2=0$.Substitute $y=(x+2)t$,then $\frac{dy}{dx}=(x+2) \frac{dt}{dx}+t$.Substitutin

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$A, D$

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(D) The given differential equation is $(x+2)(x+2+y) \frac{dy}{dx}-y^2=0$.Substitute $y=(x+2)t$,then $\frac{dy}{dx}=(x+2) \frac{dt}{dx}+t$.Substituting this into the equation: $(x+2)(x+2+(x+2)t) \frac{dy}{dx}-((x+2)t)^2=0$.$(x+2)^2(1+t) \frac{dy}{dx}-(x+2)^2t^2=0$.$(1+t)((x+2) \frac{dt}{dx}+t)-t^2=0$.$(1+t)(x+2) \frac{dt}{dx}+t+t^2-t^2=0$.$(1+t)(x+2) \frac{dt}{dx}+t=0$.Separating variables: $\frac{1+t}{t} dt = -\frac{dx}{x+2}$.Integrating both sides: $\int (\frac{1}{t}+1) dt = -\int \frac{dx}{x+2}$.$\ln|t|+t = -\ln|x+2|+C$.Substituting $t=\frac{y}{x+2}$: $\ln(\frac{y}{x+2})+\frac{y}{x+2} = -\ln(x+2)+C$.$\ln y - \ln(x+2) + \frac{y}{x+2} = -\ln(x+2)+C$.$\ln y + \frac{y}{x+2} = C$.Since the curve passes through $(1,3)$,$\ln 3 + \frac{3}{1+2} = C \Rightarrow \ln 3 + 1 = C$.The solution curve is $\ln y + \frac{y}{x+2} = \ln 3 + 1$.For option $(A)$,intersect $y=x+2$: $\ln(x+2) + \frac{x+2}{x+2} = \ln 3 + 1 \Rightarrow \ln(x+2) + 1 = \ln 3 + 1 \Rightarrow x+2=3 \Rightarrow x=1$. Only one point.For option $(D)$,$y=(x+3)^2$: $\ln(x+3)^2 + \frac{(x+3)^2}{x+2} = \ln 3 + 1$. Let $f(x) = 2\ln(x+3) + \frac{(x+3)^2}{x+2} - \ln 3 - 1$. Since $f(x)$ is strictly increasing for $x>0$ and $f(0) > 0$,it does not intersect.

$A$ solution curve of the differential equation $(x^2+xy+4x+2y+4) \frac{dy}{dx}-y^2=0, x>0$,passes through the point $(1,3)$. Then the solution curve

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