Solve the initial value problem?

[Success]

Solve the initial value problem y'=2t(1+y), y(0)=0 by the method of successive approximations.

I don't know how to do this problem but I think there's integral involved in it. Please help me. Thanks.

 

[Mandelbroth]

Solve the initial value problem y'=2t(1+y), y(0)=0 by the method of successive approximations.

I don't know how to do this problem but I think there's integral involved in it. Please help me. Thanks.

Does it have to be done by the method of successive approximations? I mean, it looks fairly simple to just say
$$\frac{dy}{dt}=2t+2ty \\ \frac{dy}{dt}-2ty=2t \implies \frac{d\mu}{dt}=-2t\mu \implies \mu=c_1e^{-t^2} \\ \frac{d}{dt}\left[c_1e^{-t^2}y\right]=2c_1te^{-t^2} \\ y = 2e^{t^2}\int te^{-t^2}dt \\ y = 2c_2e^{t^2}-1 \\ 0 = 2c_2-1 \implies c_2=\frac{1}{2} \\ y(t)=e^{t^2}-1.$$

 

[Infrared]

Does it have to be done by the method of successive approximations? I mean, it looks fairly simple to just say
$$\frac{dy}{dt}=2t+2ty \\ \frac{dy}{dt}-2ty=2t \implies \frac{d\mu}{dt}=-2t\mu \implies \mu=c_1e^{-t^2} \\ \frac{d}{dt}\left[c_1e^{-t^2}y\right]=2c_1te^{-t^2} \\ y = 2e^{t^2}\int te^{-t^2}dt \\ y = 2c_2e^{t^2}-1 \\ 0 = 2c_2-1 \implies c_2=\frac{1}{2} \\ y(t)=e^{t^2}-1.$$

Isn't the equation separable if you wanted to solve it analytically?

$$\frac{dy}{dt}=2t(1+y) \\ \frac{dy}{1+y}=2t dt \\ \ln|1+y|=t^2+C \\ y+1=\pm e^C e^{t^2}$$ and using the initial condition gives $$y=e^{t^2}-1$$

 

[Mandelbroth]

Isn't the equation separable if you wanted to solve it analytically?

$$\frac{dy}{dt}=2t(1+y) \\ \frac{dy}{1+y}=2t dt \\ \ln|1+y|=t^2+C \\ y+1=\pm e^C e^{t^2}$$ and using the initial condition gives $$y=e^{t^2}-1$$

...But it's more fun to do it the long way!

I need to start seeing when there's an easier way to solve differential equations.

As a side note, $\frac{d}{dx}\ln{x}=\frac{1}{x}\implies\int\frac{dx}{x}=\ln{x}+C$. No need for absolute values unless you are adamant about ignoring complex numbers.

Alright. Let's do it the OP's way. Suppose we have a solution $y$. Then, for $\frac{dy}{dt}=f(t,y(t))=2t(1+y)$, we have that $\displaystyle y=\int\limits_{[t_0,t]}f(x,y(x)) \, dx$. We thus approach by the method of approximations. Define $\displaystyle y_1=\int\limits_{[t_0,t]}f(x,0) \, dx = \int\limits_{[t_0,t]}2x\, dx = t^2-t_0^2$ and $\displaystyle y_n=\int\limits_{[t_0,t]}f(x,y_{n-1}) \, dx$.

Is everything clear so far? How might you continue this approach, Success?

 

[Mandelbroth]

[...]
Suppose we have a solution $y$. Then, for $\frac{dy}{dt}=f(t,y(t))=2t(1+y)$, we have that $\displaystyle y=\int\limits_{[t_0,t]}f(x,y(x)) \, dx$. We thus approach by the method of approximations. Define $\displaystyle y_1=\int\limits_{[t_0,t]}f(x,0) \, dx = \int\limits_{[t_0,t]}2x\, dx = t^2-t_0^2$ and $\displaystyle y_n=\int\limits_{[t_0,t]}f(x,y_{n-1}) \, dx$.

Is everything clear so far? How might you continue this approach, Success?

Seeing as it's been almost a day, I will suggest that we set $t_0=0$ and point out that $\displaystyle y_n = \sum_{k=1}^{n}\left[\frac{t^{2k}}{k!}\right]$.

 

[Success]

Mandel, what's y1 and yn then in this problem?

 

[micromass]

Sucess, you have received several hints right now. I would like to hear some kind of effort from you first before we continue.

 

[Success]

y=2 integral of t(1+y)dt from 0 to t. Now I need y1 and y2, I'll try. Give me some minutes.

 

[Success]

y=2 integral of (t+ty)dt from 0 to t. How to integrate this?

 

[HallsofIvy]

y=2 integral of (t+ty)dt from 0 to t. How to integrate this?

You don't. y is an unknown function of t. You cannot integrate an unknown function!

People have been trying to tell you to separate the y and t terms:
If dy/dt= t(1+ y) then dy/(1+ y)= t dt.
$$\int \frac{dy}{1+ y}= \int t dt$$.

 

[Mark44]

Isn't the equation separable if you wanted to solve it analytically?

...But it's more fun to do it the long way!

I need to start seeing when there's an easier way to solve differential equations.

Start with the simplest method first. In this case, it's easy to see by inspection that the equation is separable. The two factors on the right side were already separate functions of y and t, so multiplying out the two factors made it less obvious that separation could be used.

 

[D H]

All of the hints so far are completely off course because they don't address the central issue of using the method of successive approximations. This approach finds a sequence of functions that hopefully converge to the solution. It is very reminiscent of fixed point iteration.

Start with some guess, call it y₀(t), that satisfies the initial conditions. There's an obvious one here: y₀(t)=0. Next iterate by solving dy_n+1(t)/dt = 2t(1+y_n), y_n+1(0)=0. The first step is easy: Just solve dy₁(t)/dt=2t(1+0)=2t, y₁(0)=0. Then keep iterating. You should see a power series arising.

 

[Success]

Thank you guys. The one HS-Scientist did was very understandable.

 

[D H]

I suspect you won't get very much credit if you submit his solution. While it is the solution to the initial value problem, it does not use the method of successive approximations.

 

[Success]

But why do we must use a hard method when we can do it using the easier one?

 

[D H]

Because sometimes there is no clear-cut analytic solution.

 

[Mark44]

All of the hints so far are completely off course because they don't address the central issue of using the method of successive approximations.

Which is something that was nagging me in the back of my mind.

But why do we must use a hard method when we can do it using the easier one?

Aside from what D H mentioned, because that is the method requested in the problem statement (emphasis added).

Solve the initial value problem y'=2t(1+y), y(0)=0 by the method of successive approximations.

 

[D H]

As an example of how to use the method of successive approximations, consider the ODE y'=y, y(0)=1. Note that this is not the problem posed in the OP. Success, it's up to you to adapt this to the problem at hand.

What this method does is to build a sequence of functions y₀(t), y₁(t), y₂(t), etc., each step being a bit closer to the solution than the previous. The relation between successive approximations is y_n'(t) = y_n-1(t) such that y_n(0)=1. The starting point is a constant function that satisfies the initial condition. In this example, y(0)=1, so y₀(t) = 1. This gives the conditions for the next approximation: y₁'(t)=1, y₁(0)=1. The solution to this IVP is y₁(t)=1+t. The next iteration yields y₂(t)=1+t+t²/2, the next y₃(t)=1+t+t²/2+t³/6.

Notice how each step builds toward toward the infinite series $y(t)=\sum_{n=0}^{\infty} \frac{t^n}{n!}$