# 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

## 2 Answers

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.

I will answer the first part of your question, i.e.,

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.

You can work in proving rigorously that the RHS of part (a) is Lipschitz continuous while the RHS of part (b) is not. See that we did not need to solve the equations, since the question did not ask for the solutions, but only about their existence and unicity.