Question

Show that the problem \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{\alpha}, \mathrm{y}(0)=0\) has at least two solutions for \(0<\alpha<1\) and one solution for \(\alpha=1\)

Short answer

For 0 < α < 1, there are at least two solutions: one nontrivial and one trivial. For α = 1, there is exactly one solution, which is the trivial solution.

Step-by-step solution

Separate variables and integrate

First, separate the variables by rewriting the differential equation as \(\frac{\mathrm{dy}}{\mathrm{y}^{\alpha}}=\mathrm{dx}\). Now, integrate both sides with respect to their corresponding variables: \(\int\frac{\mathrm{d}\mathrm{y}}{\mathrm{y}^{\alpha}}=\int\mathrm{dx}\) Evaluate the integrals to get: \(\frac{\mathrm{y}^{1 - \alpha}}{1 - \alpha}= x + C\)

Solve for y

Next, find the explicit expression for y. We have: \(\mathrm{y}(x) =[(1 - \alpha)(x + C)]^{(1 / (1 - \alpha))}\)

Determine the constant using the initial condition

Since \(\mathrm{y}(0) = 0\), we can determine the constant \(C\): \(0=[(1 - \alpha)(0+C)]^{(1 / (1 - \alpha))}\) Therefore, \(C=0\). This gives us the explicit expression for y: \(\mathrm{y}(x) = [(1 - \alpha)x]^{(1 / (1 - \alpha))}\)

Determine the number of solutions for different values of \(\alpha\)

Now, let's examine the cases \(0<\alpha<1\) and \(\alpha=1\) separately to determine the number of solutions. 1. \(0<\alpha<1\): For \(0<\alpha<1\), we have two possible solutions. One solution is \(\mathrm{y}(x) = [(1 - \alpha)x]^{(1 / (1 - \alpha))}\). The other solution is the trivial solution \(\mathrm{y}(x) = 0\). 2. \(\alpha=1\): For \(\alpha=1\), the solution becomes undefined, since it involves division by zero. Therefore, the only solution for this case is the trivial solution \(\mathrm{y}(x) = 0\). In conclusion, for the given problem, there are: 1. at least two solutions for \(0<\alpha<1\) (one nontrivial and one trivial) 2. exactly one solution (trivial) for \(\alpha=1\)