Question

Tenho 2 equação diferencial para responder, alguém sabe.....xy=4y , y(1)=3..........e tenho essa equação diferencial exata 2xydx+(x²-1)dy=0

Answer 1

$x*y'=4y\\\\x* \frac{dy}{dx}= 4y\\\\ \frac{dy}{dx} = 4y* \frac{1}{x} \\\\ \frac{dy}{4y}= \frac{dx}{x} \\\\\int \frac{dy}{4y}=\int \frac{dx}{x} \\\\ \frac{ln(y)}{4} = ln(x)+C\\\\ln(y)=4ln(x)+C\\\\ln(y)=ln(x^4)+C\\\\y=e^{ln(x^4)+C}\\\\y=e^{ln(x^4)}*e^{C}\\\\\boxed{\boxed{y=x^4*C}}$

e enunciado diz que y(1) =3
$3=1^4*C\\\\3=C\\\\\\\boxed{\boxed{y(x)=3x^4}}$

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b) 
$2xydx+(x^2-1)dy=0\\\\ (x^2-1)dy = -2xydx \\\\ \frac{dy}{y}=-2 \frac{x}{(x^2-1)} dx\\\\ \int \frac{dy}{y}=\int \frac{-2x}{(x^2-1)} dx \\\\\boxed{\boxed{ ln(y)=\int \frac{-2x}{(x^2-1)} dx }}$

resolvendo a integral fazendo substituição
u = x²-1
du = 2x dx

temos 
$\int \frac{-2x}{(x^2-1)} dx = \int \frac{-du}{u} = -ln(u)+C=-ln(x^2-1)+C$

temos:
$ln(y)= -ln(x^2-1)+C\\\\ln(y)=ln[(x^2-1)^{-1}]+C\\\\ y= e^{ln[(x^2-1)^{-1}]+C}\\\\y= (x^2-1)^{-1}*C\\\\\boxed{\boxed{y(x)= \frac{C}{(x^2-1)} }}$

Answer 2

La primera es por separación de variables

$xy'=4y\\ \\ \dfrac{dy}{y}=\dfrac{4dx}{x}\\ \\ \ln |y|=4\ln|x|+C\\ \\ y=Kx^4\\ \\ 3=K\\ \\ \\ \boxed{y=3x^4}$

La segunda...

$2xydx+(x^2-1)dy=0\\ \\ \\ \displaystyle f_x(x,y)=2xy\to f(x,y)=\int 2xy \,dx + \phi(y)\to f(x,y)=x^2y+ \phi(y)\\ \\ \\ f_y(x,y)=x^2-1\\ \\ x^2+\phi'(y)=x^2-1\\ \\ \phi'(y)=-1\\ \\ \phi(y)=-y\\ \\ \\ f(x,y)=x^2y-y\\ \\ \\ \text{Soluci\'on:}\\ \\ x^2y-y=C\\ \\ \boxed{y=\dfrac{C}{x^2-1}}$