# First-order nonlinear differential equation

bennyh
Homework Statement:
first order non linear equation
Relevant Equations:
dT/dt=a-bT-Z[1/(1+vt)^2]-uT^4

a,b,z,v,u are constant
t0=0 , T=T0
Homework Statement: first order non linear equation
Homework Equations: dT/dt=a-bT-Z[1/(1+vt)^2]-uT^4

a,b,z,v,u are constant
t0=0 , T=T0

Hi,
i need find an experession of T as function of t from this first order nonlinear equation:

dT/dt=a-bT-Z[1/(1+vt)^2]-uT^4

a,b,z,v,u are constant
t0=0 , T=T0

i don't know how to solve this equation , tanks for helpers :)

Mentor
Homework Statement: first order non linear equation
Homework Equations: dT/dt=a-bT-Z[1/(1+vt)^2]-uT^4

a,b,z,v,u are constant
t0=0 , T=T0

Homework Statement: first order non linear equation
Homework Equations: dT/dt=a-bT-Z[1/(1+vt)^2]-uT^4

a,b,z,v,u are constant
t0=0 , T=T0

Hi,
i need find an experession of T as function of t from this first order nonlinear equation:

dT/dt=a-bT-Z[1/(1+vt)^2]-uT^4

a,b,z,v,u are constant
t0=0 , T=T0

i don't know how to solve this equation , tanks for helpers :)
Welcome to the PF.

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bennyh
thank you for your response , i tried solve it with Bernoulli and Riccati Equations but some how it doesn't look normal to me due to the Riccati equation (that more general) have simple form of ##y'+py=fy^n##
bact to my equation :
##dT/dt-bT=a-Z[1/(1+vt)^2]-uT^4 ##
y=f(x) -> P(x)=b (constant num) and f(x)=-u (constant) and n=4
but i don't know how to treat a-Z[1/(1+vt)^2] in the formula.

need help :)

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bennyh
Fred Wright
I have the following suggestions. You have (rearranging terms and where primes on variables indicate differentiation w.r.t time),$$T' + bT=-uT^4 +f'(t)$$ where $$f'(t) = a- \frac{Z}{(1+vt)^2}$$ Divide both sides of the equation by ##T^4## and rearrange to get,$$T^{-4}(T'-f'(t)) + bT^{-3}=-u$$ Now make the substitution$$v(t)= \frac{-T^{-3}}{3} - f(t) \\ v'(t)=T^{-4} - f'(t) \\T^{-3}= -3v(t)-3f(t)$$We now have, after rearrangement, $$v'(t) -3bv(t)=-u + 3f(t)$$ which is in a form to apply an integration factor as outlined here,http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx. With,$$p(t)=-3b \\g(t)=-u + 3f(t)$$ from Paul's above discussion on integration factors we have$$v(t)=\frac{\int e^{-3bt}(-u +3f(t))dt + C}{e^{-3bt}} \\T=\frac{1}{(-3v(t) -3f(t))^{\frac{1}{3}}}$$

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Fred Wright
Please disregard my above post. I was totally wrong. The substitution I suggested is wrong. I feel like a fool and I apologize.

bennyh
bennyh
Please disregard my above post. I was totally wrong. The substitution I suggested is wrong. I feel like a fool and I apologize.
thanks god that you wrong cause i don't understand it :) .
Do you have any other suggestion for this equation how to solve it?

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