Solving differential equation below

Question

Does anyone know how to solve this DiffQ?

$y^{-1/3} \cdot y'+ y^{2/3} = x$

Answer 1

$$y^{-1/3} \cdot y'+ y^{2/3} = x$$ Multiply by $y^{1/3}$ $$ y'+ y =xy^{1/3} $$

This is Bernoulli's equation.

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$$y^{-1/3} \cdot y'+ y^{2/3} = x$$ Another solution. Substitute $u=y^{2/3 } \implies y^{-1/3}y'=\frac 32 u'$.

The equation becomes linear: $$\frac 32 u'+u=x$$ And solve. $$ u'+\frac 23u=\frac 23 x$$ Multiply both sides by integrating factor $\mu (x)=e^{2x/3}$ $$(ue^{2x/3})'=e^{2x/3}\frac {2x}3 $$ Integrate both sides.

Answer 2

$$y'y^{-1/3}+ y^{2/3} = x\tag{1}$$ multiply by $y^{1/3}$ $$y'+y=xy^{1/3}$$

Its a Bernoulli equation of the form

$$y'+a(x)y=b(x)y^m$$

where $a(x)$ and $b(x)$ are continuous functions and $m=1/3$. Convert to a linear differential equation by the change of variable

$$z=y^{1-m}=y^{2/3} $$

Differentiating both sides, we obtain: $$z'=\frac{2}{3}y^{-1/3}y' \implies y'y^{-1/3}=\frac{3}{2}z'$$ therefore $(1)$ becomes

$$\frac{3}{2}z'+ z = x \implies z'+\frac{2}{3}z=\frac{2}{3}x\tag{2}$$

which is a linear first order differential equation of the form

$$z'+P(x)z=Q(x)$$

the integrating factor is found by

$$\mu(x)=\text{exp}\left(\int P(x)\,dx\right)=\text{exp} \left(\int \frac{2}{3}\,dx\right)=\text{exp} \left(\frac{2x}{3}\right)$$

therefore $(2)$ becomes

$$\frac{d}{dx}\left(e^{2x/3}z\right)=e^{2x/3}\frac{2}{3}x$$

Integrate and solve for $z$. To find $y$, perform the change of variable $y=z^{-2/3}$.

Answer 3

Given the equation

$y^{-1/3} \cdot y'+ y^{2/3} = x, \tag 1$

we note that

$(y^{2/3})' = \dfrac{2}{3}y^{-1/3}y', \tag 2$

whence

$\dfrac{3}{2}(y^{2/3})' = y^{-1/3}y'; \tag 3$

thus (1) may be written

$\dfrac{3}{2}(y^{2/3})' + y^{2/3} = x; \tag 4$

we set

$z = y^{2/3}; \tag 5$

and obtain

$\dfrac{3}{2}z' + z = x, \tag 6$

or

$z' + \dfrac{2}{3}z = \dfrac{2}{3}x; \tag 7$

we now multiply this equation through by $e^{2x/3}$:

$e^{2x/3}z' + \dfrac{2}{3}e^{2x/3}z = \dfrac{2}{3}e^{2x/3}x, \tag 8$

and observe that

$(e^{2x/3}z)' = e^{2x/3}z' + \dfrac{2}{3}e^{2x/3}z; \tag 9$

thus (8) becomes

$(e^{2x/3}z)' = \dfrac{2}{3}e^{2x/3}x; \tag{10}$

if we supply an initial condition $y(x_0)$ at $x_0$, then

$z(x_0) = y^{2/3}(x_0), \tag{11}$

and we may integrate (10) 'twixt $x_0$ and $x$:

$e^{2x/3}z(x) - e^{2x_0/3}z(x_0)$ $= \displaystyle \int_{x_0}^x (e^{2s/3}z(s))' \; ds = \dfrac{2}{3} \int_{x_0}^x e^{2s/3}s \; ds, \tag{12}$

and the right-hand side is now relatively easy to evaluate; $z(x)$ is then obtained via a multiplication by $e^{-2x/3}$.

**************** To Wit: ************

$(e^{2s/3}s)' = \dfrac{2}{3}e^{2s/3}s + e^{2s/3}, \tag{13}$

whence

$\dfrac{2}{3}e^{2s/3}s = (e^{2s/3}s)' - e^{2s/3}; \tag{14}$

thus we have

$\displaystyle \dfrac{2}{3} \int_{x_0}^x e^{2s/3}s \; ds = \int_{x_0}^x (e^{2s/3}s)' \; ds - \int_{x_0}^x e^{2s/3} \; ds \tag{15}$

now,

$\displaystyle \int_{x_0}^x (e^{2s/3}s)' \; ds = e^{2x/3}x - e^{2x_0/3}x_0, \tag{16}$

and

$\displaystyle \int_{x_0}^x e^{2s/3} \; ds = \dfrac{3}{2}e^{2x/3} - \dfrac{3}{2}e^{2x_0/3}; \tag{17}$

these may be inserted into (15):

$\displaystyle \dfrac{2}{3} \int_{x_0}^x e^{2s/3}s \; ds$ $= e^{2x/3}x - e^{2x_0/3}x_0 - \dfrac{3}{2}e^{2x/3} + \dfrac{3}{2}e^{2x_0/3},\tag{18}$

and hence (12) becomes

$e^{2x/3}z(x) - e^{2x_0/3}z(x_0)$ $= e^{2x/3}x - e^{2x_0/3}x_0 - \dfrac{3}{2}e^{2x/3} + \dfrac{3}{2}e^{2x_0/3},\tag{19}$

or

$e^{2x/3}z(x) = e^{2x_0/3}z(x_0)$ $+ e^{2x/3}x - e^{2x_0/3}x_0 - \dfrac{3}{2}e^{2x/3} + \dfrac{3}{2}e^{2x_0/3},\tag{20}$

which when multiplied through by $e^{-2x/3}$ becomes

$z(x) = e^{2(x_0 - x)/3}z(x_0)$ $+ x - e^{(2x_0 - x)/3}x_0 - \dfrac{3}{2} + \dfrac{3}{2}e^{2(x_0 - x)/3}; \tag{21}$

in accord with (5) and (11):

$y^{2/3}(x) = e^{2(x_0 - x)/3}y^{2/3}(x_0)$ $+ x - e^{(2x_0 - x)/3}x_0 - \dfrac{3}{2} + \dfrac{3}{2}e^{2(x_0 - x)/3}. \tag{21}$