Let y=y(x) be the solution of the differentia | vedclass

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(B) The given differential equation is $\frac{dy}{dx} = \frac{y(4y^2+2x^2)}{x(3y^2+x^2)}$.Substituting $y=vx$,we get $\frac{dy}{dx} = v + x\frac{dv}{d

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$3$

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(B) The given differential equation is $\frac{dy}{dx} = \frac{y(4y^2+2x^2)}{x(3y^2+x^2)}$.Substituting $y=vx$,we get $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv}{dx} = \frac{vx(4v^2x^2+2x^2)}{x(3v^2x^2+x^2)} = \frac{v(4v^2+2)}{3v^2+1}$.$x\frac{dv}{dx} = \frac{4v^3+2v}{3v^2+1} - v = \frac{4v^3+2v-3v^3-v}{3v^2+1} = \frac{v^3+v}{3v^2+1}$.Separating variables: $\int \frac{3v^2+1}{v^3+v} dv = \int \frac{dx}{x}$.Integrating both sides: $\ln|v^3+v| = \ln|x| + C$.Substituting $v = \frac{y}{x}$: $\ln|\frac{y^3}{x^3} + \frac{y}{x}| = \ln|x| + C$.Using $y(1)=1$: $\ln|1+1| = \ln(1) + C \Rightarrow C = \ln 2$.So,$\ln|\frac{y^3+yx^2}{x^3}| = \ln(2x)$.For $x=2$: $\frac{y^3+4y}{8} = 2(2) = 4 \Rightarrow y^3+4y = 32$.Let $f(y) = y^3+4y-32$. Since $f(2) = 8+8-32 = -16$ and $f(3) = 27+12-32 = 7$,the root $y(2)$ lies between $2$ and $3$.Thus,$y(2) \in [2, 3)$,which means $n=3$.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=\frac{4y^3+2yx^2}{3xy^2+x^3}$ with $y(1)=1$. If for some $n \in N$,$y(2) \in [n-1, n)$,then $n$ is equal to...

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