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So I just got back into differential equations recently and I'm having trouble with quite a few things: I'll give an example and explain what I don't understand:

Let the Cauchy problem $\dfrac{dy}{dx} = 3(y-1)^{2/3}$ with $y(0) = 1.$

Step 1: separating variables, we have: $\dfrac{dy}{3(y-1)^{2/3}} = dx$

Step 2 (That step I don't understand) therefore $(y-1)^{1/3} = x+c$

(How do we know that and why is it $(y-1)^{1/3}$ and not $(y-1)^{2/3} $

Step 3 (That step I don't understand either) using the initial condition we have $y(x) = x^3 + 1$

(Where does the ^3 come from?)

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  • $\begingroup$ For completeness, note that this differential equation is a classical example of nonuniqueness of the solutions since every $y_c$ for $c\geqslant0$ solves it, where $y_c(x)=1$ if $x<c$ and $y(x)=1+(x-c)^3$ for $x\geqslant c$. As a consequence, the method to solve this differential question that you present is flawed since it gives only the solution $y_0$. $\endgroup$
    – Did
    Mar 22 '17 at 14:01
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Step 2:

Let $u = y-1$. Then $du = dy$ and we have \begin{align*} \int \frac{dy}{3 (y-1)^{2/3}} &= \int \frac{du}{3u^{2/3}}\\[0.3cm] &= \frac13 \int u^{-2/3} \, du\\[0.3cm] &= \frac13 \cdot \frac1{-\frac23 + 1} u^{-\frac23 + 1} + C\\[0.3cm] &= u^{1/3} + C\\[0.3cm] &= (y-1)^{1/3} + C \end{align*}

Step 3:

We have $(y-1)^{1/3} = x + C$. We also know that when $x = 0$, we get $y = 1$. Therefore $(1 - 1)^{1/3} = 0 + C$, which means $C = 0$. Therefore $(y-1)^{1/3} = x$. Solve for $y$ to get $y = x^3 + 1$.

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Step 2: the derivative of $(y-1)^{1/3}$ with respect to $y$ is $\dfrac{1}{3(y-1)^{2/3}}$.

Step 3: From $(y(x)-1)^{1/3} = x+c$ we get with $y(0)=1$: $c=0$.

Hence $(y(x)-1)^{1/3} = x$ , thus $y(x)-1=x^3$, which gives $y(x)=x^3+1$

Remark: there is a second solution of the Cauchy problem:

$$y(x)=1$$

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  • $\begingroup$ "Remark: there is a second solution" Remark: there are tons of other solutions. $\endgroup$
    – Did
    Mar 22 '17 at 14:02

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