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how much does 2635626673925328273478277462x835382638373537283625272

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Bucky3518

18.11.2021

Mathematics

$\dfrac{dy}{dx} = 4y^2 x^{1/3}$ answer: the general solution is $y = -\dfrac{1}{3x^{\frac{4}{3} } + c}$ (note that y = 0 is also a solution because d/dx(0) = 0 and substituting y = 0 results in the given differential equation being zero.)explanation: this is a separable differential equation as y and x are factors of the derivative.we separate the y terms with the x terms, treating dy/dx as a fraction of two differentials. $\begin{aligned} \dfrac{dy}{dx} & = 4y^2 x^{1/3} \\ \dfrac{dy}{y^2} & = 4x^{1/3} \, dx \\ y^{-2} \,dy & = 4x^{1/3} \, dx \end{aligned}$ integrate both sides and then solve for y $\begin{aligned} \int y^{-2} \,dy & = \int 4x^{\frac{1}{3} } \, dx \\ \frac{y^{-1}}{-1} & = 4 \cdot \frac{x^{\frac{4}{3} } }{\frac{4}{3}} + c & & (\text{\footnotesize antiderivative power rule}) \\ -\frac{1}{y} & = 4 \cdot \tfrac{3}{4} \cdot x^{\frac{4}{3} } + c & & \left({\text{\footnotesize division by fraction is the same}}\atop{\text{\footnotesize as multiplying by reciprocal}}\right)\\-\frac{1}{y} & = 3x^{\frac{4}{3} } + c\\-1& =y\left( 3x^{\frac{4}{3}}+c\right)\\y& =-\frac{1}{3x^{\frac{4}{3}}+c}\end{aligned}$

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