# Find general solutions (implicit if necessary, explicit if convenient) of

Find general solutions (implicit if necessary, explicit if convenient) of the differential equations in Problem. Primes denote derivatives with respect to x.
$\frac{dy}{dx}={\left(64xy\right)}^{\frac{1}{3}}$
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deginasiba
$\frac{dy}{{y}^{\frac{1}{3}}}={\left(64\right)}^{\frac{1}{3}}\left({x}^{\frac{1}{3}}\right)dx$
$\int \frac{dy}{{y}^{\frac{1}{3}}}=\int 4\left({x}^{\frac{1}{3}}\right)dx$
$\frac{3{y}^{\frac{2}{3}}}{2}=\frac{3\cdot 4{x}^{\frac{4}{3}}}{4}+C$
$\frac{3{y}^{\frac{2}{3}}}{2}=3{x}^{\frac{4}{3}}+C$
${y}^{\frac{2}{3}}=2{x}^{\frac{4}{3}}+\frac{2}{3}C$
$y={\left(2{x}^{\frac{4}{3}}+\frac{2}{3}C\right)}^{\frac{3}{2}}$
Here, C is the constant of integration,
Suppose $\frac{2}{3}$ is equal to another constant D
Rewrite the solution as
$y={\left(2{x}^{\frac{4}{3}}+D\right)}^{\frac{3}{2}}$ which is the general explicit solution the given differential equation.
###### Not exactly what you’re looking for?
temnimam2
$x-\left(\frac{7}{9}\right)=\frac{13}{9}$
$\frac{67}{100}=0$
$4\cdot {\left(\mathrm{ln}\left(x\right)\right)}^{3}$
$p=\frac{w}{f}$
$\frac{6.5}{12}=0$
###### Not exactly what you’re looking for?
karton

$\frac{dy}{dx}=\left(64xy{\right)}^{\frac{1}{3}}=4{x}^{\frac{1}{3}}{y}^{\frac{1}{3}}$
or, ${y}^{-\frac{1}{3}}dy=4{x}^{\frac{1}{3}}dx$
Integrating, we get,
$\int {y}^{-\frac{1}{3}}dy=4\int {x}^{\frac{1}{3}}dx$
or, $\frac{{y}^{-\frac{1}{3}}+1}{-\frac{1}{3}+1}=4\frac{{x}^{\frac{1}{3}}+1}{\frac{1}{3}+1}+c$
or, $\frac{{y}^{\frac{2}{3}}}{\frac{2}{3}}=4\frac{{x}^{\frac{4}{3}}}{\frac{4}{3}}+c$
or, $\frac{3}{2}{y}^{2/3}=3{x}^{4/3}+c$
which is the required general solution, where c be an arbitrary constant.
$\left(1+x\right)\frac{dy}{dx}=4y$
or, $\frac{dy}{y}=4\frac{dx}{x+1}$
Integrating, $\int \frac{dy}{y}=4\int \frac{dx}{x+1}$
or, $\mathrm{ln}|y|=4\mathrm{ln}|x+1|+\mathrm{ln}c$
or, $y=c\left(x+1{\right)}^{4}$
Required general solution:
$y=c\left(x+1{\right)}^{4}$ where c be an arbitrary constant.