Solving second order ordinary differential equation with variable constants
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I'm having trouble solving a differential equation I found:
$$ u''(x) + xint_0^xu(t)dt = f(x) $$
where: $ xin[0,1], quad u(0) = 1, quad u(1) = -1 $, and $f(x)$ any given function.
One of my problems is I don't really understand the term of the integral because with that it doesn't look like a linear second order ordinary differential equation or anything I have ever seen before. What I have tried is converting it to a system of differential equations and also tried to solve it as a homogenous equation, but it didn't turned out into anything useful,
If anyone can help me getting started I would be very grateful.
differential-equations computational-mathematics integral-equations
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
asked Nov 23 at 17:34
hugdelcur96
165
add a comment |
up vote
0
down vote
favorite
I'm having trouble solving a differential equation I found:
$$ u''(x) + xint_0^xu(t)dt = f(x) $$
where: $ xin[0,1], quad u(0) = 1, quad u(1) = -1 $, and $f(x)$ any given function.
One of my problems is I don't really understand the term of the integral because with that it doesn't look like a linear second order ordinary differential equation or anything I have ever seen before. What I have tried is converting it to a system of differential equations and also tried to solve it as a homogenous equation, but it didn't turned out into anything useful,
If anyone can help me getting started I would be very grateful.
differential-equations computational-mathematics integral-equations
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
asked Nov 23 at 17:34
hugdelcur96
165
1
Let $v(x) = int_0^x u(t) , mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$
– MisterRiemann
Nov 23 at 17:57
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having trouble solving a differential equation I found:
$$ u''(x) + xint_0^xu(t)dt = f(x) $$
where: $ xin[0,1], quad u(0) = 1, quad u(1) = -1 $, and $f(x)$ any given function.
One of my problems is I don't really understand the term of the integral because with that it doesn't look like a linear second order ordinary differential equation or anything I have ever seen before. What I have tried is converting it to a system of differential equations and also tried to solve it as a homogenous equation, but it didn't turned out into anything useful,
If anyone can help me getting started I would be very grateful.
differential-equations computational-mathematics integral-equations
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
asked Nov 23 at 17:34
hugdelcur96
165
I'm having trouble solving a differential equation I found:
$$ u''(x) + xint_0^xu(t)dt = f(x) $$
where: $ xin[0,1], quad u(0) = 1, quad u(1) = -1 $, and $f(x)$ any given function.
One of my problems is I don't really understand the term of the integral because with that it doesn't look like a linear second order ordinary differential equation or anything I have ever seen before. What I have tried is converting it to a system of differential equations and also tried to solve it as a homogenous equation, but it didn't turned out into anything useful,
If anyone can help me getting started I would be very grateful.
differential-equations computational-mathematics integral-equations
differential-equations computational-mathematics integral-equations
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
asked Nov 23 at 17:34
hugdelcur96
165
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
asked Nov 23 at 17:34
hugdelcur96
165
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
edited Nov 24 at 12:40
doraemonpaul
12.5k31660
12.5k31660
asked Nov 23 at 17:34
hugdelcur96
165
asked Nov 23 at 17:34
hugdelcur96
165
asked Nov 23 at 17:34
hugdelcur96
165
165
1
Let $v(x) = int_0^x u(t) , mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$
– MisterRiemann
Nov 23 at 17:57
add a comment |
1
Let $v(x) = int_0^x u(t) , mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$
– MisterRiemann
Nov 23 at 17:57
1
1
Let $v(x) = int_0^x u(t) , mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$
– MisterRiemann
Nov 23 at 17:57
Let $v(x) = int_0^x u(t) , mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$
– MisterRiemann
Nov 23 at 17:57
add a comment |
1 Answer
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With $U(x)=int_0^x u(t),dt$ you get $U'=u$, $U(0)=0$ and inserted it gives a third order ODE
$$
U'''(x)+xU(x)=f(x)
$$
with initial conditions $U(0)=0$, $U'(0)=u(0)=1$ and $U'(1)=u(1)=-1$.
answered Nov 23 at 17:57
LutzL
55.3k42053
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
up vote
2
down vote
With $U(x)=int_0^x u(t),dt$ you get $U'=u$, $U(0)=0$ and inserted it gives a third order ODE
$$
U'''(x)+xU(x)=f(x)
$$
with initial conditions $U(0)=0$, $U'(0)=u(0)=1$ and $U'(1)=u(1)=-1$.
answered Nov 23 at 17:57
LutzL
55.3k42053
add a comment |
up vote
2
down vote
With $U(x)=int_0^x u(t),dt$ you get $U'=u$, $U(0)=0$ and inserted it gives a third order ODE
$$
U'''(x)+xU(x)=f(x)
$$
with initial conditions $U(0)=0$, $U'(0)=u(0)=1$ and $U'(1)=u(1)=-1$.
answered Nov 23 at 17:57
LutzL
55.3k42053
add a comment |
up vote
2
down vote
up vote
2
down vote
With $U(x)=int_0^x u(t),dt$ you get $U'=u$, $U(0)=0$ and inserted it gives a third order ODE
$$
U'''(x)+xU(x)=f(x)
$$
with initial conditions $U(0)=0$, $U'(0)=u(0)=1$ and $U'(1)=u(1)=-1$.
answered Nov 23 at 17:57
LutzL
55.3k42053
With $U(x)=int_0^x u(t),dt$ you get $U'=u$, $U(0)=0$ and inserted it gives a third order ODE
$$
U'''(x)+xU(x)=f(x)
$$
with initial conditions $U(0)=0$, $U'(0)=u(0)=1$ and $U'(1)=u(1)=-1$.
answered Nov 23 at 17:57
LutzL
55.3k42053
answered Nov 23 at 17:57
LutzL
55.3k42053
answered Nov 23 at 17:57
LutzL
55.3k42053
answered Nov 23 at 17:57
LutzL
55.3k42053
55.3k42053
add a comment |
add a comment |
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Let $v(x) = int_0^x u(t) , mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$
– MisterRiemann
Nov 23 at 17:57