Question

The differential equation dy/dx=3y^2/3, y(0)=0 has a unique solution.
True or false with full explanation

Answer 1

Given differential equation: $\frac{dy}{dx}=3y^{\frac{2}{3}}$

The above differential equation is of the form

$\frac{dy}{dx}=F(x,y)$ , where $F(x,y)=3y^{\frac{2}{3}}$

Observe that both $F(x,y)$ and $\frac{\partial F(x,y)}{\partial y}$ are continuous functions on $(-\infty ,\infty )$ and

it contains 0.

Therefore,, the given initial value problem has unique solution by using Existence and uniqueness theorem.

To find the solution, use separation of variables.

Using separation of variables, we get

$\frac{dy}{y^{\frac{2}{3}}}=3dx$

Integrating on both sides, we get

$\int \frac{dy}{y^{\frac{2}{3}}}=3\int dx$

$\frac{y^{\frac{1}{3}}}{\frac{1}{3}}=3x+C$

That is, we get

$3y^{\frac{1}{3}}=3x+C$

Use the initial condition $y(0)=0$ to find $C$

Now $y(0)=0\Rightarrow 3(0)+C=0\Rightarrow C=0$

Therefore, solution to the initial value problem is

$3y^{\frac{1}{3}}=3x$ or $y^{\frac{1}{3}}=x$