What's the General Solution to This IVP?

logan3
Messages
83
Reaction score
2
\frac {dy}{dt} = y^3 + t^2, y(0) = 0

My teacher said this IVP couldn't be expressed in terms of functions we commonly know. I was wondering what the general solution is?

Thank-you
 
Physics news on Phys.org
logan3 said:
\frac {dy}{dt} = y^3 + t^2, y(0) = 0

My teacher said this IVP couldn't be expressed in terms of functions we commonly know. I was wondering what the general solution is?

Thank-you
For solving this
It is a bernoulli form
Watch this video and you will get it how to solve
 
##\frac {dy}{dt} = y^3 + t^2, y(0) = 0## is not of the form ##y' + f(t)y = g(t)y^n## so it is not the Bernoulli equation.
 
Yes it's not Bernoulli form, sorry. I not know how to solve it, but Wolframalpha was showing the simple answer x=1 for y=0.
I don't know how it solved that.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
5
Views
2K
Replies
6
Views
2K
Replies
8
Views
4K
Replies
6
Views
2K
Replies
3
Views
2K
Replies
1
Views
3K
Replies
5
Views
2K
Replies
8
Views
2K
Replies
5
Views
5K
Back
Top