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In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem dy/dx=f(x,y), y(x) = yo. If f and f/y are continuous functions in some rectangle R= {(x,y):a Use the method of separation of variables to find the solution to dy/dx =y1/3? Begin by separating the variables. dy= dx

Respuesta :

Answer:

The answer is "[tex]\bold{ y^{ \frac{2}{3}}} = \frac{2}{3}x + c} \\\\[/tex]"

Step-by-step explanation:

By using the given method we  separate the above-given variable and  find the solution of:

[tex]\bold{\frac{dy}{dx} =y^{\frac{1}{3}}}[/tex]

Solution:

[tex]\to \frac{dy}{dx} =y^{\frac{1}{3}}\\\\\to \frac{dy}{y^{\frac{1}{3}}} =dx\\\\\to y^{ - \frac{1}{3}} dy =dx\\\\[/tex]

Integrate the above value:

[tex]\to \int y^{ - \frac{1}{3}} dy = \int 1 dx\\\\\\\to \frac{y^{ - \frac{1}{3} +1}}{- \frac{1}{3} +1} = x \\\\ \\\to \frac{y^{ \frac{-1 +3}{3}}}{ \frac{-1+3}{3} +1} = x \\\\\\\to \frac{y^{ \frac{2}{3}}}{ \frac{2}{3} +1} = x \\\\\\\to \bold{y^{ \frac{2}{3}}} = \frac{2}{3}x + c} \\\\[/tex]

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