# what is the solution of this ODE

How to solve the following differential equation

$$\frac{dy}{dx} + \frac{5y}{6x} = \frac{5x^4}{y^4}$$

subject to the condition $y(1) = 1$.

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–  Raskolnikov Sep 16 '12 at 18:00
Take a look at the Wolfram|Alpha solution : bit.ly/UbkhQf –  gyosko Sep 16 '12 at 18:35

Generally, a Bernoulli’s equation is as:

$$\frac{dy}{dx}+P(x)y=f(x)y^n,$$

where $n$ is any real number. It can be reduced to a linear equation by the substitution $w=y^{1-n}$. Here, $n=-4$, $f(x)=5x^4$ and $P(x)=5/(6x)$. Your equation is reduced to:

$$\frac{dw}{dx}+5\left(\frac{5}{6x}\right)w=5(5x^4),$$

which is linear equation. Solving it you get:

$$w = \frac{30}{11}x^5+\frac{C}{x^{25/6}}.$$

Now put $w=y^5$ and satisfy your initial condition.

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@enzotib: Thanks for editting. :) –  Babak S. Sep 17 '12 at 11:32
$\ddot \smile \quad +1\quad$ –  amWhy Mar 18 '13 at 0:59
@amWhy: I don't know what would happen to me if you were not be here as a friend. ;-) –  Babak S. Mar 18 '13 at 2:39
thank you, dear! Your friendship (and participation here) makes math.SE all the brighter! –  amWhy Mar 18 '13 at 2:40

Hint:

This is equivalent to $5x^{a}y'y^4+ay^5x^{a-1}=b(a+5)x^{a+4}$, for some suitable $a$ and $b$. The LHS is the derivative of $x^{a}y^5$ hence $x^ay^5=bx^{a+5}+c$, for some constant $c$. Take it from here.

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This looks like a Bernoulli differential equation with $n=-4$ so let's use the change of variable $w:=y^5$ (i.e. $w:=1/y^{n-1}$) to get $w'=5y^4y'$ that we will replace in your equation :

$$y^4y'+\frac {5y^5}{6x}=5x^4$$ $$\frac {w'}5+\frac {5w}{6x}=5x^4$$

But $\frac {w'}5+\frac {5w}{6x}=0\$ is $\ \frac {w'}w=-\frac {25}6\frac 1x$ with the solution $\log w= -\frac {25}6 \log(x)+C_0\$ that is :
$$w=C\cdot x^{-25/6}$$ We may for example use variation of constants to get : $$w'=C'\cdot x^{-25/6}-\frac {25\,C}6x^{-31/6}$$ and solve the complete ODE : $$\frac {w'}5+\frac {5w}{6x}=\frac {C'}5\cdot x^{-25/6}=5x^4$$ so that $C'=25\,x^{4+25/6}=25\,x^{49/6}$ that gives $\ C=\frac {30}{11}x^{\frac{55}6}+C_1$ and : $$w=\left(\frac {30}{11}x^{\frac{55}6}+C_1\right)\cdot x^{-25/6}$$ $$w=\frac {30}{11}x^5+C_1\, x^{-25/6}$$ That you may replace in $w=y^5$ to conclude (using your initial condition $w(1)=1$).

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I don't know if it's better or worse, but I've recently found another way to solve problems of this sort. The first step is to proceed as if it were a linear ODE by finding an integrating factor.

$$x^{\frac56}\frac{dy}{dx}+\frac56x^{-\frac16}y=\frac{5x^{\frac{29}{6}}}{y^4}$$

The left hand side is now $\frac d{dx}(x^{\frac56}y)$. Accordingly, we'll make the substitution

$$z=x^{\frac56}y,y=x^{-\frac56}z$$

$$\frac{dz}{dx}=\frac{5x^\frac{29}6}{x^{-\frac{20}6}z^4}=5x^\frac{49}6z^{-4}$$

Our equation is now separable.

$$z^4dz=5x^\frac{49}6dx$$

$$\frac15z^5=5(\frac6{55})x^\frac{55}6+k_1$$

$$z^5=\frac{30}{11}x^\frac{55}6+k_2$$

$$x^\frac{25}6y^5=\frac{30}{11}x^\frac{55}6+k_2$$

$$y^5=\frac{30}{11}x^5+k_2x^{-\frac{25}6}$$

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