solve the equation $uu'+2e^u-t+e^ut=0$, $u(0)=1$
1
$begingroup$
I am trying to solve the equation, $uu'+2e^u-t+e^ut=0$, with $u(0)=1$.
This problem is giving me trouble because I don't know how to get the equation into the form of $dy/dx=g(x)f(y)$.
calculus
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
$endgroup$
add a comment |
1
$begingroup$
I am trying to solve the equation, $uu'+2e^u-t+e^ut=0$, with $u(0)=1$.
This problem is giving me trouble because I don't know how to get the equation into the form of $dy/dx=g(x)f(y)$.
calculus
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
$endgroup$
$begingroup$
Doesn't u' = du/dt?
$endgroup$
– William Elliot
Dec 9 '18 at 10:21
$begingroup$
I think so, but I'm at a loss as to what to do after that.
$endgroup$
– Levi Bentley
Dec 9 '18 at 19:36
add a comment |
1
1
1
2
$begingroup$
I am trying to solve the equation, $uu'+2e^u-t+e^ut=0$, with $u(0)=1$.
This problem is giving me trouble because I don't know how to get the equation into the form of $dy/dx=g(x)f(y)$.
calculus
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
$endgroup$
I am trying to solve the equation, $uu'+2e^u-t+e^ut=0$, with $u(0)=1$.
This problem is giving me trouble because I don't know how to get the equation into the form of $dy/dx=g(x)f(y)$.
calculus
calculus
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
edited Dec 9 '18 at 4:13
Key Flex
7,94461233
7,94461233
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
asked Dec 9 '18 at 4:10
Levi BentleyLevi Bentley
61
61
$begingroup$
Doesn't u' = du/dt?
$endgroup$
– William Elliot
Dec 9 '18 at 10:21
$begingroup$
I think so, but I'm at a loss as to what to do after that.
$endgroup$
– Levi Bentley
Dec 9 '18 at 19:36
add a comment |
$begingroup$
Doesn't u' = du/dt?
$endgroup$
– William Elliot
Dec 9 '18 at 10:21
$begingroup$
I think so, but I'm at a loss as to what to do after that.
$endgroup$
– Levi Bentley
Dec 9 '18 at 19:36
$begingroup$
Doesn't u' = du/dt?
$endgroup$
– William Elliot
Dec 9 '18 at 10:21
$begingroup$
Doesn't u' = du/dt?
$endgroup$
– William Elliot
Dec 9 '18 at 10:21
$begingroup$
I think so, but I'm at a loss as to what to do after that.
$endgroup$
– Levi Bentley
Dec 9 '18 at 19:36
$begingroup$
I think so, but I'm at a loss as to what to do after that.
$endgroup$
– Levi Bentley
Dec 9 '18 at 19:36
add a comment |
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$begingroup$
Doesn't u' = du/dt?
$endgroup$
– William Elliot
Dec 9 '18 at 10:21
$begingroup$
I think so, but I'm at a loss as to what to do after that.
$endgroup$
– Levi Bentley
Dec 9 '18 at 19:36