Solve bernoulli differential equation with extra constant

[Berrius]

Homework Statement

Find a solution for:
u'(t)=c*u(t)^2-c*(a+b)*u(t)+c*a*b

The Attempt at a Solution

I've found the solution for the homogeneous equation:
u_0(t)=(\frac{1}{a+b}+d*e^{c(a+b)t)})^{-1}
Where c is a random constant.

Now I've tried the solution u(t)=x(t)*u_0(t), when I fill this in it gets a mess and i can't figure out what x(t) should be.
Can someone help me?

 

[LCKurtz]

Homework Statement

Find a solution for:
u'(t)=c*u(t)^2-c*(a+b)*u(t)+c*a*b

The Attempt at a Solution

I've found the solution for the homogeneous equation:
u_0(t)=(\frac{1}{a+b}+d*e^{c(a+b)t)})^{-1}
Where c is a random constant.

Now I've tried the solution u(t)=x(t)*u_0(t), when I fill this in it gets a mess and i can't figure out what x(t) should be.
Can someone help me?

This isn't a linear equation, so I'm not sure what you consider the "homogeneous" equation. If you write it as$$
u'+ c(a+b)u =cu^2+cab$$at least the left side of the equation is linear. That means if you can find a solution $u_1$ of $u'+ c(a+b)u =cu^2$ and a solution $u_2$ of $u'+ c(a+b)u =cab$ then $u_1+u_2$ should be a solution of your original DE.

 

[Berrius]

With homogeneous I just ment the differential equation without the constant part.

But your approach won't work because u'(t)=u_1'(t)+u_2'(t)=-c(a+b)u_1(t)+cu_1^2(t)-c(a+b)u_2(t)+cab=-c(a+b)u(t)+cu_1^2(t)+cab

Or am I seeing it wrong?

 

[Dick]

I would forget Bernoulli's equation and just try separation of variables. The right side factors. Just use partial fractions to integrate it.

 

[Dick]

With homogeneous I just ment the differential equation without the constant part.

But your approach won't work because u'(t)=u_1'(t)+u_2'(t)=-c(a+b)u_1(t)+cu_1^2(t)-c(a+b)u_2(t)+cab=-c(a+b)u(t)+cu_1^2(t)+cab

Or am I seeing it wrong?

Right. You can't combine them that way. Just separate.

 

[LCKurtz]

With homogeneous I just ment the differential equation without the constant part.

But your approach won't work because u'(t)=u_1'(t)+u_2'(t)=-c(a+b)u_1(t)+cu_1^2(t)-c(a+b)u_2(t)+cab=-c(a+b)u(t)+cu_1^2(t)+cab

Or am I seeing it wrong?

You are seeing it wrong. If you let$$
L(u) = u' + c(a+b)u$$be the left side linear part then $L$ is a linear operator, which means $L(u_1+u_2) = L(u_1)+L(u_2)$. So if $L(u_1) = cu_1^2$ and $L(u_2)=abc$, what do you get for $L(u_1+u_2)$ ?

 

[Dick]

You are seeing it wrong. If you let$$
L(u) = u' + c(a+b)u$$be the left side linear part then $L$ is a linear operator, which means $L(u_1+u_2) = L(u_1)+L(u_2)$. So if $L(u_1) = cu_1^2$ and $L(u_2)=abc$, what do you get for $L(u_1+u_2)$ ?

But you want to get $L(u_1+u_2) = c(u_1+u_2)^2+abc$. You don't get that.

 

[LCKurtz]

But you want to get $L(u_1+u_2) = c(u_1+u_2)^2+abc$. You don't get that.

No wonder I was feeling a little uncertainty about my argument. And it seemed so neat at the time...