How to do implicit differentiation to get an ODE Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Laplace question continued (partial fractions)How to differentiate the equation under the integral signFind the solution to the initial value problem $overlinex' = Aoverlinex, x(0) = beginbmatrix2 \ 28endbmatrix $How to solve this implicit differentiation problem concerning arcsin?inhomogeneous second order odeSolving a simple system of odeDetermine the Eigen values and Eigen functions of the problem $y''+μy=0$; $y(0)=0$,$y'(L)=0$Solve ODE with exactnessRearranging terms in an ODEWhat is a 'reasonable' ordinary differential equation question?
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How to do implicit differentiation to get an ODE
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Laplace question continued (partial fractions)How to differentiate the equation under the integral signFind the solution to the initial value problem $overlinex' = Aoverlinex, x(0) = beginbmatrix2 \ 28endbmatrix $How to solve this implicit differentiation problem concerning arcsin?inhomogeneous second order odeSolving a simple system of odeDetermine the Eigen values and Eigen functions of the problem $y''+μy=0$; $y(0)=0$,$y'(L)=0$Solve ODE with exactnessRearranging terms in an ODEWhat is a 'reasonable' ordinary differential equation question?
$begingroup$
$$ye^y^2 = x + q $$
Hi I was following a similar question as an example to answer this one and although I have got the answer I don't think my working is correct as I sort of made one part up and I don't fully understand it. I was hoping someone could help and check to see if my working is correct.
Line $1$ is the question and the Last Line is the correct answer but I don't get line $4$ and $5$. I just followed an example I was looking at but I don't actually understand it and I'm pretty sure it's not correct either.
Any help will be appreciated thank you.
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
$$ye^y^2 = x + q $$
Hi I was following a similar question as an example to answer this one and although I have got the answer I don't think my working is correct as I sort of made one part up and I don't fully understand it. I was hoping someone could help and check to see if my working is correct.
Line $1$ is the question and the Last Line is the correct answer but I don't get line $4$ and $5$. I just followed an example I was looking at but I don't actually understand it and I'm pretty sure it's not correct either.
Any help will be appreciated thank you.
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
$$ye^y^2 = x + q $$
Hi I was following a similar question as an example to answer this one and although I have got the answer I don't think my working is correct as I sort of made one part up and I don't fully understand it. I was hoping someone could help and check to see if my working is correct.
Line $1$ is the question and the Last Line is the correct answer but I don't get line $4$ and $5$. I just followed an example I was looking at but I don't actually understand it and I'm pretty sure it's not correct either.
Any help will be appreciated thank you.
ordinary-differential-equations
$endgroup$
$$ye^y^2 = x + q $$
Hi I was following a similar question as an example to answer this one and although I have got the answer I don't think my working is correct as I sort of made one part up and I don't fully understand it. I was hoping someone could help and check to see if my working is correct.
Line $1$ is the question and the Last Line is the correct answer but I don't get line $4$ and $5$. I just followed an example I was looking at but I don't actually understand it and I'm pretty sure it's not correct either.
Any help will be appreciated thank you.
ordinary-differential-equations
ordinary-differential-equations
edited Apr 1 at 1:45
John Doe
12.1k11339
12.1k11339
asked Apr 1 at 1:26
CinnaCinna
193
193
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The step you seem to be confused over is computing $fracddx(y^2)$. This is $$fracddxy^2=fracdydxfrac ddy(y^2)=y'times 2y=2yy'$$ Then the rest of the working follows fine.
$endgroup$
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The step you seem to be confused over is computing $fracddx(y^2)$. This is $$fracddxy^2=fracdydxfrac ddy(y^2)=y'times 2y=2yy'$$ Then the rest of the working follows fine.
$endgroup$
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
add a comment |
$begingroup$
The step you seem to be confused over is computing $fracddx(y^2)$. This is $$fracddxy^2=fracdydxfrac ddy(y^2)=y'times 2y=2yy'$$ Then the rest of the working follows fine.
$endgroup$
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
add a comment |
$begingroup$
The step you seem to be confused over is computing $fracddx(y^2)$. This is $$fracddxy^2=fracdydxfrac ddy(y^2)=y'times 2y=2yy'$$ Then the rest of the working follows fine.
$endgroup$
The step you seem to be confused over is computing $fracddx(y^2)$. This is $$fracddxy^2=fracdydxfrac ddy(y^2)=y'times 2y=2yy'$$ Then the rest of the working follows fine.
answered Apr 1 at 1:32
John DoeJohn Doe
12.1k11339
12.1k11339
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
add a comment |
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
$begingroup$
That makes a lot more sense thank you so much :)
$endgroup$
– Cinna
Apr 1 at 1:43
add a comment |
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