Question

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

Answer 1

The answer is "
`\bold{ y^{ (2)/(3)}} = (2)/(3)x + c} \\\\`"

Explanation:

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

`\bold{(dy)/(dx) =y^{(1)/(3)}}`

Solution:

`\to (dy)/(dx) =y^{(1)/(3)}\\\\\to \frac{dy}{y^{(1)/(3)}} =dx\\\\\to y^{ - (1)/(3)} dy =dx\\\\`

Integrate the above value:

`\to \int y^{ - (1)/(3)} dy = \int 1 dx\\\\\\\to \frac{y^{ - (1)/(3) +1}}{- (1)/(3) +1} = x \\\\ \\\to \frac{y^{ (-1 +3)/(3)}}{ (-1+3)/(3) +1} = x \\\\\\\to \frac{y^{ (2)/(3)}}{ (2)/(3) +1} = x \\\\\\\to \bold{y^{ (2)/(3)}} = (2)/(3)x + c} \\\\`