Determining solution to Differential Equation with given initial value

Prove that there is exactly one solution of the differential equation $$(a)$$ $$y'=5y^{\frac{5}{4}}$$ with initial data $$y(0) = y_0$$. Is this statement true for the equation $$(b)$$ $$y'=5y^{\frac{4}{5}}$$?

I calculated the $$1^{st}$$ differential equation and obtained as a solution

$$y=\frac{256y_0^{\frac{1}{4}}}{(-5ty_0^{\frac{1}{4}}-4)^4}.$$

Is this correct? Or is it equal to $$y=(-\frac{5t}{4}+y_0^\frac{-1}{4})^{-4}$$?

I got two different answers, as you could see, the first one I got by dividing the constant I got after integration by $$-4$$ and just went on to applying the initial conditions, yet for the second answer I just left it as c, then applied the initial condition. I believe the $$2^{nd}$$ answer may be correct. Any help would be greatly appreciated.

Also, for the second part I have as a solution that $$y=t^5 + y_0$$, which I believe I did correctly, again, any help would be greatly appreciated. Also, how would I know that there is exactly one solution for $$(a)$$, and if the statement is true for $$(b)$$?

• I think that you are supposed to use the Picard-Lindelöf theorem. – rafa11111 Dec 13 '18 at 11:38
• Questions end with a question mark. – Klangen Dec 13 '18 at 11:48

In the first equation you get $$(y(t)^{-\frac14})'=-\frac14y(t)^{-\frac54}y'(t)=-\frac54 \\ \implies y(t)^{-\frac14}-y_0^{-\frac14}=-\frac54t \\ \implies y(t)=\frac{y_0}{\left(1-\frac54y_0^{\frac14}t\right)^4}$$ which is equivalent to your second answer for $$y_0>0$$. In your first answer, you missed to transfer some signs and exponents correctly.

The second equation can be solved similar to that, you should get $$y(t)=\left(y_0^{1/5}+t\right)^5,$$ I'm not sure how you got to your result. The only similarity is that for $$y_0=0$$ you get the solution $$y(t)=t^5$$. But you should see that for that initial value you get another trivial solution.

Prove that there is exactly one solution of the differential equation $$(a)$$ $$y'=5y^{\frac{5}{4}}$$ with initial data $$y(0) = y_0$$. Is this statement true for the equation $$(b)$$ $$y'=5y^{\frac{4}{5}}$$?
From the Picard-Lindelöf theorem, the solution of the differential equation $$y'=f(t,y)$$, with initial condition $$y(t_0)=y_0$$ exists and is unique if $$f(t,y)$$ is Lipschitz continuous in $$y$$ and continuous in $$t$$.
In the case (a), $$f(t,y)=5y^{5/4}$$, that is Lipschitz continuous if there is $$K$$ such that $$|5y_1^{5/4}-5y_2^{5/4}|\leq K |y_1-y_2|.$$ However, instead of using the definition, we can use the fact that a continuous function is Lipschitz continuous if its derivative is limited and the fact that $$5y^{5/4}$$ is continuous in $$y\geq0$$. Its derivative is $$25/4 y^{1/4}$$, which is limited in any interval you choose. Therefore, since $$5y^{5/4}$$ is Lipschitz continuous, there is only one solution to the differential equation $$y'=5y^{5/4}$$.
Repeating the same argument for the part (b), we can use the fact that $$5y^{4/5}$$ is not Lipschitz continuous for all $$y\geq0$$, since its derivative, $$4y^{-1/5}$$, is infinite as $$y\to 0$$. Therefore, the same holds for part (b) if $$y>0$$. Since the RHS of the equation is always positive, $$y(t)$$ grows monotonically. If $$y_0>0$$, there is only one solution for (b). If $$y_0=0$$, however, it's not true.