Question

Determine a solução de cada EDO, utilizando o método da separação de variáveis e sua solução particular: a) y' = 3x²y / y(0)=1 b) y' = -4xy² / y(0) = 4 c) y'-2x = 0 / y(2) = d) x.dy-y².dx = 0 / y(1) = 3 e) x.dy - y².dx = 0 / y(0) = 6 f) dy/dx = x^2y^4 / y(0) = 10 PASSO A PASSO POR FAVOR.

Answer 1

Olá,


Equações Diferenciais Ordinárias por Variáveis Separáveis.


A)

$\displaystyle \mathsf{y'=3x^2y\qquad\qquad\qquad y(0)=1}\\\\\\\\\mathsf{ \frac{dy}{dx}=3x^2y }\\\\\\\mathsf{ \frac{dy}{y}=3x^2dx }\\\\\\\text{Integra dos dois lados}\\\\\\\mathsf{ \int \frac{dy}{y}=\int 3x^2dx }\\\\\\\mathsf{\ell n |y|= \frac{\diagup\!\!\!\!3x^3 }{\diagup\!\!\!\!3} +C }\\\\\\\mathsf{\ell n |y|= x^3 +C }$

y(0) = 1
x = 0      e     y = 1

$\displaystyle \mathsf{\underbrace{\ell n |1|}_{=0}= 0^3 +C }\\\\\\\mathsf{C=0}\\\\\\\\\mathsf{\ell n |y|= x^3}\\\\\\\\\text{Deixando o 'y' de forma explicita}\\\\\\\text{Aplica exponencia nos dois lados, para eliminar o }\ell n\\\\\\\mathsf{e^{\ell n |y|}=e^{x^3}}\\\\\\\boxed{\mathsf{y=e^{x^3}}}$



B)

$\displaystyle \mathsf{y'=-4xy^2\qquad\qquad\qquad y(0)=4}\\\\\\\mathsf{ \frac{dy}{dx}=-4xy^2 }\\\\\\\mathsf{ \int \frac{dy}{y^2}=\int -4xdx }\\\\\\\mathsf{- \frac{1}{y}=-2x^2+C }$

y(0) = 4
x = 0   e   y = 4

$\displaystyle\mathsf{- \frac{1}{4}=-2(0)^2+C }\\\\\\\mathsf{C=- \frac{1}{4} }\\\\\\\mathsf{- \frac{1}{y}=-2x^2- \frac{1}{4} }\\\\\\\mathsf{-1=\left(-2x^2- \frac{1}{4} \right)y}\\\\\\\mathsf{y= \frac{-1}{-2x^2- \frac{1}{4}} }\\\\\\\mathsf{y= \frac{-1}{- \frac{8x^2-1}{4} } }\\\\\\\boxed{\mathsf{y= \frac{4}{-8x^2-1} }}$



C)

$\displaystyle \mathsf{y'-2x=0\qquad\qquad\qquad y(2)=0}\\\\\\ \mathsf{\frac{dy}{dx}=2x }\\\\\\\mathsf{\int dy=\int 2xdx}\\\\\\\mathsf{y=x^2+C}\\\\\\\text{x=2, y=0}\\\\\\\mathsf{0=2^2+C}\\\\\\\mathsf{C=-4}\\\\\\\boxed{\mathsf{y= x^2-4}}$


D)

$\displaystyle \mathsf{xdy-y^2dx=0\qquad\qquad\qquad y(1)=3}\\\\\\\mathsf{xdy=y^2dx}\\\\\\\mathsf{ \int \frac{dy}{y^2}=\int \frac{dx}{x} }\\\\\\\mathsf{- \frac{1}{y}=\ell n|x|+C }\\\\\\\text{x=1,y=3}\\\\\\\mathsf{- \frac{1}{3}= \ell n|1|+C}\\\\\\\mathsf{C=- \frac{1}{3} }\\\\\\\mathsf{- \frac{1}{y}=\ell n|x|- \frac{1}{3} }\\\\\\\mathsf{-1=\left(\ell n|x|- \frac{1}{3} \right)y}\\\\\\\mathsf{y= \frac{-1}{\ell n|x|- \frac{1}{3} } }\\\\\\\mathsf{y= \frac{-1}{ \frac{3\ell n |x|-1}{3} } }$

$\displaystyle \boxed{\mathsf{y= \frac{-3}{3\ell n |x|-1 } }}$


E)

Alguma coisa no valor inicial está errado.

$\displaystyle \mathsf{xdy-y^2dx=0\qquad\qquad\qquad y(0)=6}\\\\\\\mathsf{ \int \frac{dy}{y^2}=\int \frac{dx}{x} }\\\\\\\mathsf{- \frac{1}{y}=\ell n|x|+C }\\\\\\\text{x=0,y=6}\\\\\\\mathsf{\ell n (0) ~nao ~existe, ~verifique~ a~ questao}$


F)

$\displaystyle \mathsf{ \frac{dy}{dx}=x^2y^4\qquad\qquad\qquad y( 0)=10}\\\\\\\mathsf{ \int \frac{dy}{y^4}=\int x^2dx }\\\\\\\\\mathsf{- \frac{1}{3y^3}= \frac{x^3}{3}+C }\\\\\\\text{x=0,y=10}\\\\\\\\\mathsf{- \frac{1}{3\cdot (10)^3}= \frac{0^3}{3}+C }\\\\\\\mathsf{C=- \frac{1}{3000} }\\\\\\\mathsf{- \frac{1}{3y^3}= \frac{x^3}{3} - \frac{1}{3000} }\\\\\\\mathsf{- \frac{1}{y^3}=x^3- \frac{1}{1000} }\\\\\\\mathsf{-1=\left(x^3- \frac{1}{1000} \right)y^3}\\\\\\\mathsf{y^3= \frac{-1}{x^3- \frac{1}{1000} } }$

$\displaystyle \mathsf{y^3= \frac{-1}{x^3- \frac{1}{1000} } }\\\\\\\mathsf{y^3= \frac{-1}{ \frac{1000x^3-1}{1000} } }\\\\\\\mathsf{y^3= \frac{-1000}{1000x^3-1} }\\\\\\\boxed{\mathsf{y= \sqrt[3]{\frac{-1000}{1000x^3-1} } }}$