Setting linear coefficient of the Riccati equation to one
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Consider a Ricatti equation $u'=au^2+bu=0$. i.e. Riccati equation without free term and constant coefficients $a$, $b$.. Is it possible to make a linear transformation such that $b$ becomes 1? I want to find a linear transformation such that above equation becomes $u'=au^2+u$. So far, I have that it can't be done using linear transformations.
ordinary-differential-equations derivatives
asked Dec 7 '18 at 23:07
SoleSole
61
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add a comment |
$begingroup$
Consider a Ricatti equation $u'=au^2+bu=0$. i.e. Riccati equation without free term and constant coefficients $a$, $b$.. Is it possible to make a linear transformation such that $b$ becomes 1? I want to find a linear transformation such that above equation becomes $u'=au^2+u$. So far, I have that it can't be done using linear transformations.
ordinary-differential-equations derivatives
asked Dec 7 '18 at 23:07
SoleSole
61
$endgroup$
1
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You can change $b$ by rescaling the time variable. But note that this is also a Bernoulli equation, formulating it in terms of $u^{-1}$ simplifies it immediately.
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– LutzL
Dec 7 '18 at 23:37
1
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Using the change $u=v−1$, original equation in terms of v became $−v′=bv+a$. New equation has no quadratic term and can be solved analytically as a linear ODE. Thank you very much!
$endgroup$
– Sole
Dec 8 '18 at 0:25
add a comment |
$begingroup$
Consider a Ricatti equation $u'=au^2+bu=0$. i.e. Riccati equation without free term and constant coefficients $a$, $b$.. Is it possible to make a linear transformation such that $b$ becomes 1? I want to find a linear transformation such that above equation becomes $u'=au^2+u$. So far, I have that it can't be done using linear transformations.
ordinary-differential-equations derivatives
asked Dec 7 '18 at 23:07
SoleSole
61
$endgroup$
Consider a Ricatti equation $u'=au^2+bu=0$. i.e. Riccati equation without free term and constant coefficients $a$, $b$.. Is it possible to make a linear transformation such that $b$ becomes 1? I want to find a linear transformation such that above equation becomes $u'=au^2+u$. So far, I have that it can't be done using linear transformations.
ordinary-differential-equations derivatives
ordinary-differential-equations derivatives
asked Dec 7 '18 at 23:07
SoleSole
61
asked Dec 7 '18 at 23:07
SoleSole
61
asked Dec 7 '18 at 23:07
SoleSole
61
asked Dec 7 '18 at 23:07
SoleSole
61
asked Dec 7 '18 at 23:07
SoleSole
61
61
1
$begingroup$
You can change $b$ by rescaling the time variable. But note that this is also a Bernoulli equation, formulating it in terms of $u^{-1}$ simplifies it immediately.
$endgroup$
– LutzL
Dec 7 '18 at 23:37
1
$begingroup$
Using the change $u=v−1$, original equation in terms of v became $−v′=bv+a$. New equation has no quadratic term and can be solved analytically as a linear ODE. Thank you very much!
$endgroup$
– Sole
Dec 8 '18 at 0:25
add a comment |
1
$begingroup$
You can change $b$ by rescaling the time variable. But note that this is also a Bernoulli equation, formulating it in terms of $u^{-1}$ simplifies it immediately.
$endgroup$
– LutzL
Dec 7 '18 at 23:37
1
$begingroup$
Using the change $u=v−1$, original equation in terms of v became $−v′=bv+a$. New equation has no quadratic term and can be solved analytically as a linear ODE. Thank you very much!
$endgroup$
– Sole
Dec 8 '18 at 0:25
1
1
$begingroup$
You can change $b$ by rescaling the time variable. But note that this is also a Bernoulli equation, formulating it in terms of $u^{-1}$ simplifies it immediately.
$endgroup$
– LutzL
Dec 7 '18 at 23:37
$begingroup$
You can change $b$ by rescaling the time variable. But note that this is also a Bernoulli equation, formulating it in terms of $u^{-1}$ simplifies it immediately.
$endgroup$
– LutzL
Dec 7 '18 at 23:37
1
1
$begingroup$
Using the change $u=v−1$, original equation in terms of v became $−v′=bv+a$. New equation has no quadratic term and can be solved analytically as a linear ODE. Thank you very much!
$endgroup$
– Sole
Dec 8 '18 at 0:25
$begingroup$
Using the change $u=v−1$, original equation in terms of v became $−v′=bv+a$. New equation has no quadratic term and can be solved analytically as a linear ODE. Thank you very much!
$endgroup$
– Sole
Dec 8 '18 at 0:25
add a comment |
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1
$begingroup$
You can change $b$ by rescaling the time variable. But note that this is also a Bernoulli equation, formulating it in terms of $u^{-1}$ simplifies it immediately.
$endgroup$
– LutzL
Dec 7 '18 at 23:37
1
$begingroup$
Using the change $u=v−1$, original equation in terms of v became $−v′=bv+a$. New equation has no quadratic term and can be solved analytically as a linear ODE. Thank you very much!
$endgroup$
– Sole
Dec 8 '18 at 0:25