Finding a primitive function of a partial derivative of a function by integrating
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Assume I have an expression e.g. $dfrac{partial}{partial x}left(dfrac{partial u}{partial y}right)=0, ,$ : $, ,u=u(x,y)$.
How would I proceed (in a correct way) finding a general expression of $u(x,y)$ by integrating?
calculus algebra-precalculus
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$begingroup$
Assume I have an expression e.g. $dfrac{partial}{partial x}left(dfrac{partial u}{partial y}right)=0, ,$ : $, ,u=u(x,y)$.
How would I proceed (in a correct way) finding a general expression of $u(x,y)$ by integrating?
calculus algebra-precalculus
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add a comment |
$begingroup$
Assume I have an expression e.g. $dfrac{partial}{partial x}left(dfrac{partial u}{partial y}right)=0, ,$ : $, ,u=u(x,y)$.
How would I proceed (in a correct way) finding a general expression of $u(x,y)$ by integrating?
calculus algebra-precalculus
$endgroup$
Assume I have an expression e.g. $dfrac{partial}{partial x}left(dfrac{partial u}{partial y}right)=0, ,$ : $, ,u=u(x,y)$.
How would I proceed (in a correct way) finding a general expression of $u(x,y)$ by integrating?
calculus algebra-precalculus
calculus algebra-precalculus
asked Feb 1 at 14:28
user615771
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2 Answers
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Firstly
$$
frac{partial u}{partial x} = C_1 + phi(y)
$$
and secondly
$$
u(x,y) = C_1 y + C_2 + int_0^yphi(tau) dtau + psi(x)
$$
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$begingroup$
Integrating once wrt $x$, you get
$$ dfrac{partial u}{partial y} = f_1(y) $$
where $f_1(y)$ is an arbitrary function of y only.
Then integrating this wrt $y$, you get an antiderivative of $f_1$ (and since $f_1$ is arbitrary, so is its antiderivative: call it $g(y)$) plus a "constant", which in this case means an arbitrary function of $x$. The final result is:
$$ u(x, y) = f(x) + g(y) $$
where $f$ and $g$ are arbitrary functions.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
Firstly
$$
frac{partial u}{partial x} = C_1 + phi(y)
$$
and secondly
$$
u(x,y) = C_1 y + C_2 + int_0^yphi(tau) dtau + psi(x)
$$
$endgroup$
add a comment |
$begingroup$
Firstly
$$
frac{partial u}{partial x} = C_1 + phi(y)
$$
and secondly
$$
u(x,y) = C_1 y + C_2 + int_0^yphi(tau) dtau + psi(x)
$$
$endgroup$
add a comment |
$begingroup$
Firstly
$$
frac{partial u}{partial x} = C_1 + phi(y)
$$
and secondly
$$
u(x,y) = C_1 y + C_2 + int_0^yphi(tau) dtau + psi(x)
$$
$endgroup$
Firstly
$$
frac{partial u}{partial x} = C_1 + phi(y)
$$
and secondly
$$
u(x,y) = C_1 y + C_2 + int_0^yphi(tau) dtau + psi(x)
$$
answered Feb 1 at 16:28
CesareoCesareo
9,8063517
9,8063517
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$begingroup$
Integrating once wrt $x$, you get
$$ dfrac{partial u}{partial y} = f_1(y) $$
where $f_1(y)$ is an arbitrary function of y only.
Then integrating this wrt $y$, you get an antiderivative of $f_1$ (and since $f_1$ is arbitrary, so is its antiderivative: call it $g(y)$) plus a "constant", which in this case means an arbitrary function of $x$. The final result is:
$$ u(x, y) = f(x) + g(y) $$
where $f$ and $g$ are arbitrary functions.
$endgroup$
add a comment |
$begingroup$
Integrating once wrt $x$, you get
$$ dfrac{partial u}{partial y} = f_1(y) $$
where $f_1(y)$ is an arbitrary function of y only.
Then integrating this wrt $y$, you get an antiderivative of $f_1$ (and since $f_1$ is arbitrary, so is its antiderivative: call it $g(y)$) plus a "constant", which in this case means an arbitrary function of $x$. The final result is:
$$ u(x, y) = f(x) + g(y) $$
where $f$ and $g$ are arbitrary functions.
$endgroup$
add a comment |
$begingroup$
Integrating once wrt $x$, you get
$$ dfrac{partial u}{partial y} = f_1(y) $$
where $f_1(y)$ is an arbitrary function of y only.
Then integrating this wrt $y$, you get an antiderivative of $f_1$ (and since $f_1$ is arbitrary, so is its antiderivative: call it $g(y)$) plus a "constant", which in this case means an arbitrary function of $x$. The final result is:
$$ u(x, y) = f(x) + g(y) $$
where $f$ and $g$ are arbitrary functions.
$endgroup$
Integrating once wrt $x$, you get
$$ dfrac{partial u}{partial y} = f_1(y) $$
where $f_1(y)$ is an arbitrary function of y only.
Then integrating this wrt $y$, you get an antiderivative of $f_1$ (and since $f_1$ is arbitrary, so is its antiderivative: call it $g(y)$) plus a "constant", which in this case means an arbitrary function of $x$. The final result is:
$$ u(x, y) = f(x) + g(y) $$
where $f$ and $g$ are arbitrary functions.
answered Feb 1 at 20:08
NickDNickD
1,2171512
1,2171512
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