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How to solve this differential equation? and what type is it?
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I would like to know what kind of differential equation is this? A linear ODE, a nonlinear DE, etc? And also what will be the ways of solving it?
$$
fracdT(t)dt + AsqrtT(t)cdot(T(t)-B) + Ccdot (T(t)+D)^4 = E + F(GT(t)+H)
$$
I have seen in the books that the equation is solved as an ODE by linearising the $T^4(t)$ term and making it separable.
Thanks
differential-equations pde
asked Jul 30 at 9:38
user401393
84
 |Â
show 7 more comments
up vote
0
down vote
favorite
I would like to know what kind of differential equation is this? A linear ODE, a nonlinear DE, etc? And also what will be the ways of solving it?
$$
fracdT(t)dt + AsqrtT(t)cdot(T(t)-B) + Ccdot (T(t)+D)^4 = E + F(GT(t)+H)
$$
I have seen in the books that the equation is solved as an ODE by linearising the $T^4(t)$ term and making it separable.
Thanks
differential-equations pde
asked Jul 30 at 9:38
user401393
84
Are you sure that $f_1$, etc., are functions of $T$, not of $t$? Because then, if we do not know what $f_1$ etc. are, there are no general methods of finding a solution in closed form (only abstract theorems on the existence (and possibly uniqueness) of an initial value problem).
– user539887
Jul 30 at 9:44
So if $f_1$ = $1+2T^2(t)+3T^3(t)$, should I say it is a function of $f_1(T)$ or $f_1(t)$?
– user401393
Jul 30 at 9:50
@user401393 As a function of $T$.
– xbh
Jul 30 at 9:52
Such an $f_1$ is a function of $T$.
– user539887
Jul 30 at 9:52
2
No, not at all linear; it is a separable (autonomous) equation, of the form $T'(t)=g(T(t))$. If you need a closed form, the problem is in finding the inverse function of $intfracdTg(T)$.
– user539887
Jul 30 at 11:15
 |Â
show 7 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to know what kind of differential equation is this? A linear ODE, a nonlinear DE, etc? And also what will be the ways of solving it?
$$
fracdT(t)dt + AsqrtT(t)cdot(T(t)-B) + Ccdot (T(t)+D)^4 = E + F(GT(t)+H)
$$
I have seen in the books that the equation is solved as an ODE by linearising the $T^4(t)$ term and making it separable.
Thanks
differential-equations pde
asked Jul 30 at 9:38
user401393
84
I would like to know what kind of differential equation is this? A linear ODE, a nonlinear DE, etc? And also what will be the ways of solving it?
$$
fracdT(t)dt + AsqrtT(t)cdot(T(t)-B) + Ccdot (T(t)+D)^4 = E + F(GT(t)+H)
$$
I have seen in the books that the equation is solved as an ODE by linearising the $T^4(t)$ term and making it separable.
Thanks
differential-equations pde
asked Jul 30 at 9:38
user401393
84
edited Jul 30 at 10:51
asked Jul 30 at 9:38
user401393
84
asked Jul 30 at 9:38
user401393
84
asked Jul 30 at 9:38
user401393
84
84
Are you sure that $f_1$, etc., are functions of $T$, not of $t$? Because then, if we do not know what $f_1$ etc. are, there are no general methods of finding a solution in closed form (only abstract theorems on the existence (and possibly uniqueness) of an initial value problem).
– user539887
Jul 30 at 9:44
So if $f_1$ = $1+2T^2(t)+3T^3(t)$, should I say it is a function of $f_1(T)$ or $f_1(t)$?
– user401393
Jul 30 at 9:50
@user401393 As a function of $T$.
– xbh
Jul 30 at 9:52
Such an $f_1$ is a function of $T$.
– user539887
Jul 30 at 9:52
2
No, not at all linear; it is a separable (autonomous) equation, of the form $T'(t)=g(T(t))$. If you need a closed form, the problem is in finding the inverse function of $intfracdTg(T)$.
– user539887
Jul 30 at 11:15
 |Â
show 7 more comments
Are you sure that $f_1$, etc., are functions of $T$, not of $t$? Because then, if we do not know what $f_1$ etc. are, there are no general methods of finding a solution in closed form (only abstract theorems on the existence (and possibly uniqueness) of an initial value problem).
– user539887
Jul 30 at 9:44
So if $f_1$ = $1+2T^2(t)+3T^3(t)$, should I say it is a function of $f_1(T)$ or $f_1(t)$?
– user401393
Jul 30 at 9:50
@user401393 As a function of $T$.
– xbh
Jul 30 at 9:52
Such an $f_1$ is a function of $T$.
– user539887
Jul 30 at 9:52
2
No, not at all linear; it is a separable (autonomous) equation, of the form $T'(t)=g(T(t))$. If you need a closed form, the problem is in finding the inverse function of $intfracdTg(T)$.
– user539887
Jul 30 at 11:15
Are you sure that $f_1$, etc., are functions of $T$, not of $t$? Because then, if we do not know what $f_1$ etc. are, there are no general methods of finding a solution in closed form (only abstract theorems on the existence (and possibly uniqueness) of an initial value problem).
– user539887
Jul 30 at 9:44
Are you sure that $f_1$, etc., are functions of $T$, not of $t$? Because then, if we do not know what $f_1$ etc. are, there are no general methods of finding a solution in closed form (only abstract theorems on the existence (and possibly uniqueness) of an initial value problem).
– user539887
Jul 30 at 9:44
So if $f_1$ = $1+2T^2(t)+3T^3(t)$, should I say it is a function of $f_1(T)$ or $f_1(t)$?
– user401393
Jul 30 at 9:50
So if $f_1$ = $1+2T^2(t)+3T^3(t)$, should I say it is a function of $f_1(T)$ or $f_1(t)$?
– user401393
Jul 30 at 9:50
@user401393 As a function of $T$.
– xbh
Jul 30 at 9:52
@user401393 As a function of $T$.
– xbh
Jul 30 at 9:52
Such an $f_1$ is a function of $T$.
– user539887
Jul 30 at 9:52
Such an $f_1$ is a function of $T$.
– user539887
Jul 30 at 9:52
2
2
No, not at all linear; it is a separable (autonomous) equation, of the form $T'(t)=g(T(t))$. If you need a closed form, the problem is in finding the inverse function of $intfracdTg(T)$.
– user539887
Jul 30 at 11:15
No, not at all linear; it is a separable (autonomous) equation, of the form $T'(t)=g(T(t))$. If you need a closed form, the problem is in finding the inverse function of $intfracdTg(T)$.
– user539887
Jul 30 at 11:15
 |Â
show 7 more comments
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Are you sure that $f_1$, etc., are functions of $T$, not of $t$? Because then, if we do not know what $f_1$ etc. are, there are no general methods of finding a solution in closed form (only abstract theorems on the existence (and possibly uniqueness) of an initial value problem).
– user539887
Jul 30 at 9:44
So if $f_1$ = $1+2T^2(t)+3T^3(t)$, should I say it is a function of $f_1(T)$ or $f_1(t)$?
– user401393
Jul 30 at 9:50
@user401393 As a function of $T$.
– xbh
Jul 30 at 9:52
Such an $f_1$ is a function of $T$.
– user539887
Jul 30 at 9:52
2
No, not at all linear; it is a separable (autonomous) equation, of the form $T'(t)=g(T(t))$. If you need a closed form, the problem is in finding the inverse function of $intfracdTg(T)$.
– user539887
Jul 30 at 11:15