Mixing
Riccati ODE Solution Path
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In the Ricatti Ordinary Differential Equation introduction in this video, the form of the general solution is stated as $y = y_1 + u$ and the Ricatti ODE becomes a Bernoulli ODE, which eventually becomes linear, while in this video, the stated general solution is $y = y_1 + frac{1}{v}$, and the Ricatti ODE becomes a linear ODE directly, with no intermediate Bernoulli step. Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step? If not, why is $y = y_1 + frac{1}{v}$ sometimes preferred, and what conditions determine whether the Ricatti ODE will go through Bernoulli form as an intermediate step, or will become linear immediately?
ordinary-differential-equations
asked Dec 27 '18 at 5:08
user10478user10478
448211
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add a comment |
$begingroup$
In the Ricatti Ordinary Differential Equation introduction in this video, the form of the general solution is stated as $y = y_1 + u$ and the Ricatti ODE becomes a Bernoulli ODE, which eventually becomes linear, while in this video, the stated general solution is $y = y_1 + frac{1}{v}$, and the Ricatti ODE becomes a linear ODE directly, with no intermediate Bernoulli step. Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step? If not, why is $y = y_1 + frac{1}{v}$ sometimes preferred, and what conditions determine whether the Ricatti ODE will go through Bernoulli form as an intermediate step, or will become linear immediately?
ordinary-differential-equations
asked Dec 27 '18 at 5:08
user10478user10478
448211
$endgroup$
add a comment |
$begingroup$
In the Ricatti Ordinary Differential Equation introduction in this video, the form of the general solution is stated as $y = y_1 + u$ and the Ricatti ODE becomes a Bernoulli ODE, which eventually becomes linear, while in this video, the stated general solution is $y = y_1 + frac{1}{v}$, and the Ricatti ODE becomes a linear ODE directly, with no intermediate Bernoulli step. Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step? If not, why is $y = y_1 + frac{1}{v}$ sometimes preferred, and what conditions determine whether the Ricatti ODE will go through Bernoulli form as an intermediate step, or will become linear immediately?
ordinary-differential-equations
asked Dec 27 '18 at 5:08
user10478user10478
448211
$endgroup$
In the Ricatti Ordinary Differential Equation introduction in this video, the form of the general solution is stated as $y = y_1 + u$ and the Ricatti ODE becomes a Bernoulli ODE, which eventually becomes linear, while in this video, the stated general solution is $y = y_1 + frac{1}{v}$, and the Ricatti ODE becomes a linear ODE directly, with no intermediate Bernoulli step. Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step? If not, why is $y = y_1 + frac{1}{v}$ sometimes preferred, and what conditions determine whether the Ricatti ODE will go through Bernoulli form as an intermediate step, or will become linear immediately?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 27 '18 at 5:08
user10478user10478
448211
asked Dec 27 '18 at 5:08
user10478user10478
448211
asked Dec 27 '18 at 5:08
user10478user10478
448211
asked Dec 27 '18 at 5:08
user10478user10478
448211
asked Dec 27 '18 at 5:08
user10478user10478
448211
448211
add a comment |
add a comment |
1 Answer
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The answer to "Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step?" is yes. Why?
The Riccati equation always has the form:
$$y' = A(x) + B(x)y + C(x)y^2$$
We will write it like this:
$$y' = A + By + Cy^2$$
And (as the video says) if you use the sustitution $y = y_P(x) + u$ where $y_P(x)$ is a particular solution (and known) of the Ricatti equation, you have:
$$y'_P(x) + u' = A + B[y_P(x)+u] + C[y_P(x) + u]^2$$
$$y'_P(x) + u' = A + left[By_P(x) + Buright] + left[Cy_P(x)^2 + 2Cy_P(x)u + Cu^2right]$$
But $y_P(x)$ is solution:
$$y'_P(x) = A + By_P(x) + Cy_P(x)^2$$
Then if you substitute $y'_P(x)$ as $A + By_P(x) + Cy_P(x)^2$ in the Ricatti equation:
$$A + By_P(x) + Cy_P^2 + u' = A + By_P + Bu + Cy_P^2 + 2Cy_Pu + Cu^2$$
$$u' = Bu + 2Cy_Pu + Cu^2$$
This is a Bernoulli equation, so that we have a linear equation, we must use the substitution $displaystyle v=u^{1-2}=frac{1}{u}$. If you can notice:
$$v=frac{1}{u}=frac{1}{y-y_P(x)}$$
$$v=frac{1}{y-y_P(x)}$$
$$y-y_P(x)=frac{1}{v}$$
$$y=y_P+frac{1}{v}$$
Then if you had started with the substitution $displaystyle y=y_P+frac{1}{v}$ you would have arrived at the linear equation to have missed the Bernoulli step, but it is exactly the same.
answered Jan 6 at 8:48
El boritoEl borito
575216
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answer to "Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step?" is yes. Why?
The Riccati equation always has the form:
$$y' = A(x) + B(x)y + C(x)y^2$$
We will write it like this:
$$y' = A + By + Cy^2$$
And (as the video says) if you use the sustitution $y = y_P(x) + u$ where $y_P(x)$ is a particular solution (and known) of the Ricatti equation, you have:
$$y'_P(x) + u' = A + B[y_P(x)+u] + C[y_P(x) + u]^2$$
$$y'_P(x) + u' = A + left[By_P(x) + Buright] + left[Cy_P(x)^2 + 2Cy_P(x)u + Cu^2right]$$
But $y_P(x)$ is solution:
$$y'_P(x) = A + By_P(x) + Cy_P(x)^2$$
Then if you substitute $y'_P(x)$ as $A + By_P(x) + Cy_P(x)^2$ in the Ricatti equation:
$$A + By_P(x) + Cy_P^2 + u' = A + By_P + Bu + Cy_P^2 + 2Cy_Pu + Cu^2$$
$$u' = Bu + 2Cy_Pu + Cu^2$$
This is a Bernoulli equation, so that we have a linear equation, we must use the substitution $displaystyle v=u^{1-2}=frac{1}{u}$. If you can notice:
$$v=frac{1}{u}=frac{1}{y-y_P(x)}$$
$$v=frac{1}{y-y_P(x)}$$
$$y-y_P(x)=frac{1}{v}$$
$$y=y_P+frac{1}{v}$$
Then if you had started with the substitution $displaystyle y=y_P+frac{1}{v}$ you would have arrived at the linear equation to have missed the Bernoulli step, but it is exactly the same.
answered Jan 6 at 8:48
El boritoEl borito
575216
$endgroup$
add a comment |
$begingroup$
The answer to "Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step?" is yes. Why?
The Riccati equation always has the form:
$$y' = A(x) + B(x)y + C(x)y^2$$
We will write it like this:
$$y' = A + By + Cy^2$$
And (as the video says) if you use the sustitution $y = y_P(x) + u$ where $y_P(x)$ is a particular solution (and known) of the Ricatti equation, you have:
$$y'_P(x) + u' = A + B[y_P(x)+u] + C[y_P(x) + u]^2$$
$$y'_P(x) + u' = A + left[By_P(x) + Buright] + left[Cy_P(x)^2 + 2Cy_P(x)u + Cu^2right]$$
But $y_P(x)$ is solution:
$$y'_P(x) = A + By_P(x) + Cy_P(x)^2$$
Then if you substitute $y'_P(x)$ as $A + By_P(x) + Cy_P(x)^2$ in the Ricatti equation:
$$A + By_P(x) + Cy_P^2 + u' = A + By_P + Bu + Cy_P^2 + 2Cy_Pu + Cu^2$$
$$u' = Bu + 2Cy_Pu + Cu^2$$
This is a Bernoulli equation, so that we have a linear equation, we must use the substitution $displaystyle v=u^{1-2}=frac{1}{u}$. If you can notice:
$$v=frac{1}{u}=frac{1}{y-y_P(x)}$$
$$v=frac{1}{y-y_P(x)}$$
$$y-y_P(x)=frac{1}{v}$$
$$y=y_P+frac{1}{v}$$
Then if you had started with the substitution $displaystyle y=y_P+frac{1}{v}$ you would have arrived at the linear equation to have missed the Bernoulli step, but it is exactly the same.
answered Jan 6 at 8:48
El boritoEl borito
575216
$endgroup$
add a comment |
$begingroup$
The answer to "Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step?" is yes. Why?
The Riccati equation always has the form:
$$y' = A(x) + B(x)y + C(x)y^2$$
We will write it like this:
$$y' = A + By + Cy^2$$
And (as the video says) if you use the sustitution $y = y_P(x) + u$ where $y_P(x)$ is a particular solution (and known) of the Ricatti equation, you have:
$$y'_P(x) + u' = A + B[y_P(x)+u] + C[y_P(x) + u]^2$$
$$y'_P(x) + u' = A + left[By_P(x) + Buright] + left[Cy_P(x)^2 + 2Cy_P(x)u + Cu^2right]$$
But $y_P(x)$ is solution:
$$y'_P(x) = A + By_P(x) + Cy_P(x)^2$$
Then if you substitute $y'_P(x)$ as $A + By_P(x) + Cy_P(x)^2$ in the Ricatti equation:
$$A + By_P(x) + Cy_P^2 + u' = A + By_P + Bu + Cy_P^2 + 2Cy_Pu + Cu^2$$
$$u' = Bu + 2Cy_Pu + Cu^2$$
This is a Bernoulli equation, so that we have a linear equation, we must use the substitution $displaystyle v=u^{1-2}=frac{1}{u}$. If you can notice:
$$v=frac{1}{u}=frac{1}{y-y_P(x)}$$
$$v=frac{1}{y-y_P(x)}$$
$$y-y_P(x)=frac{1}{v}$$
$$y=y_P+frac{1}{v}$$
Then if you had started with the substitution $displaystyle y=y_P+frac{1}{v}$ you would have arrived at the linear equation to have missed the Bernoulli step, but it is exactly the same.
answered Jan 6 at 8:48
El boritoEl borito
575216
$endgroup$
The answer to "Is the difference in the assumed general solution form the reason for the presence or absence of the intermediate Bernoulli step?" is yes. Why?
The Riccati equation always has the form:
$$y' = A(x) + B(x)y + C(x)y^2$$
We will write it like this:
$$y' = A + By + Cy^2$$
And (as the video says) if you use the sustitution $y = y_P(x) + u$ where $y_P(x)$ is a particular solution (and known) of the Ricatti equation, you have:
$$y'_P(x) + u' = A + B[y_P(x)+u] + C[y_P(x) + u]^2$$
$$y'_P(x) + u' = A + left[By_P(x) + Buright] + left[Cy_P(x)^2 + 2Cy_P(x)u + Cu^2right]$$
But $y_P(x)$ is solution:
$$y'_P(x) = A + By_P(x) + Cy_P(x)^2$$
Then if you substitute $y'_P(x)$ as $A + By_P(x) + Cy_P(x)^2$ in the Ricatti equation:
$$A + By_P(x) + Cy_P^2 + u' = A + By_P + Bu + Cy_P^2 + 2Cy_Pu + Cu^2$$
$$u' = Bu + 2Cy_Pu + Cu^2$$
This is a Bernoulli equation, so that we have a linear equation, we must use the substitution $displaystyle v=u^{1-2}=frac{1}{u}$. If you can notice:
$$v=frac{1}{u}=frac{1}{y-y_P(x)}$$
$$v=frac{1}{y-y_P(x)}$$
$$y-y_P(x)=frac{1}{v}$$
$$y=y_P+frac{1}{v}$$
Then if you had started with the substitution $displaystyle y=y_P+frac{1}{v}$ you would have arrived at the linear equation to have missed the Bernoulli step, but it is exactly the same.
answered Jan 6 at 8:48
El boritoEl borito
575216
answered Jan 6 at 8:48
El boritoEl borito
575216
answered Jan 6 at 8:48
El boritoEl borito
575216
answered Jan 6 at 8:48
El boritoEl borito
575216
575216
add a comment |
add a comment |
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