Mixing
how can I find exact solution?
$begingroup$
What is the approach to solve $y'(t)=alpha(y(t)-a+bt)^2$ analytically? $alpha, a,b$ are positive constants.
I have tried MAPLE dsolve and got $y(t)=sqrt(b/ alpha) tan left[ sqrt(alpha b)(t-C) right]+a-bp$. I have checked the $y(t)$ by substitution that does confirm that it is the solution.
I wonder what method of solving or trick is used here?
ordinary-differential-equations
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
$endgroup$
add a comment |
$begingroup$
What is the approach to solve $y'(t)=alpha(y(t)-a+bt)^2$ analytically? $alpha, a,b$ are positive constants.
I have tried MAPLE dsolve and got $y(t)=sqrt(b/ alpha) tan left[ sqrt(alpha b)(t-C) right]+a-bp$. I have checked the $y(t)$ by substitution that does confirm that it is the solution.
I wonder what method of solving or trick is used here?
ordinary-differential-equations
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
$endgroup$
add a comment |
$begingroup$
What is the approach to solve $y'(t)=alpha(y(t)-a+bt)^2$ analytically? $alpha, a,b$ are positive constants.
I have tried MAPLE dsolve and got $y(t)=sqrt(b/ alpha) tan left[ sqrt(alpha b)(t-C) right]+a-bp$. I have checked the $y(t)$ by substitution that does confirm that it is the solution.
I wonder what method of solving or trick is used here?
ordinary-differential-equations
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
$endgroup$
What is the approach to solve $y'(t)=alpha(y(t)-a+bt)^2$ analytically? $alpha, a,b$ are positive constants.
I have tried MAPLE dsolve and got $y(t)=sqrt(b/ alpha) tan left[ sqrt(alpha b)(t-C) right]+a-bp$. I have checked the $y(t)$ by substitution that does confirm that it is the solution.
I wonder what method of solving or trick is used here?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
edited Jan 29 at 7:27
Zviad Khukhunashvili
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
asked Jan 29 at 6:32
Zviad KhukhunashviliZviad Khukhunashvili
1279
1279
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Substituting $$v(t)=frac{-2aalpha+2btalpha}{2alpha}+y(t)$$ then you will get
$$-b+v'(t)=alpha v(t)^2$$
or
$$frac{frac{dv(t)}{dt}}{b+alpha v(t)^2}=1$$
Can you finish?
It is a Riccati equation.
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
After change $y(t)=z+a-bt$ we get separable equation
$$z'=alpha z^2+b$$
If $alpha b<0$ MAPLE gives complex solution.
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
$endgroup$
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
add a comment |
$begingroup$
Let $y=z-a+b t$ to make the equation
$$z'-alpha z^2-b=0$$ that is to say $$frac 1 {t'}-alpha z^2-b=0implies t'=frac{1 }{ alpha z^2+b}$$ which looks to be quite simple.
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
Hint: Substituting $$v(t)=frac{-2aalpha+2btalpha}{2alpha}+y(t)$$ then you will get
$$-b+v'(t)=alpha v(t)^2$$
or
$$frac{frac{dv(t)}{dt}}{b+alpha v(t)^2}=1$$
Can you finish?
It is a Riccati equation.
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
Hint: Substituting $$v(t)=frac{-2aalpha+2btalpha}{2alpha}+y(t)$$ then you will get
$$-b+v'(t)=alpha v(t)^2$$
or
$$frac{frac{dv(t)}{dt}}{b+alpha v(t)^2}=1$$
Can you finish?
It is a Riccati equation.
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
Hint: Substituting $$v(t)=frac{-2aalpha+2btalpha}{2alpha}+y(t)$$ then you will get
$$-b+v'(t)=alpha v(t)^2$$
or
$$frac{frac{dv(t)}{dt}}{b+alpha v(t)^2}=1$$
Can you finish?
It is a Riccati equation.
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
Hint: Substituting $$v(t)=frac{-2aalpha+2btalpha}{2alpha}+y(t)$$ then you will get
$$-b+v'(t)=alpha v(t)^2$$
or
$$frac{frac{dv(t)}{dt}}{b+alpha v(t)^2}=1$$
Can you finish?
It is a Riccati equation.
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 29 at 6:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
add a comment |
add a comment |
$begingroup$
After change $y(t)=z+a-bt$ we get separable equation
$$z'=alpha z^2+b$$
If $alpha b<0$ MAPLE gives complex solution.
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
$endgroup$
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
add a comment |
$begingroup$
After change $y(t)=z+a-bt$ we get separable equation
$$z'=alpha z^2+b$$
If $alpha b<0$ MAPLE gives complex solution.
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
$endgroup$
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
add a comment |
$begingroup$
After change $y(t)=z+a-bt$ we get separable equation
$$z'=alpha z^2+b$$
If $alpha b<0$ MAPLE gives complex solution.
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
$endgroup$
After change $y(t)=z+a-bt$ we get separable equation
$$z'=alpha z^2+b$$
If $alpha b<0$ MAPLE gives complex solution.
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
answered Jan 29 at 7:18
Aleksas DomarkasAleksas Domarkas
1,63317
1,63317
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
add a comment |
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
$begingroup$
Forgot to mention that all constants are positive. My apologies. Edited.
$endgroup$
– Zviad Khukhunashvili
Jan 29 at 7:27
add a comment |
$begingroup$
Let $y=z-a+b t$ to make the equation
$$z'-alpha z^2-b=0$$ that is to say $$frac 1 {t'}-alpha z^2-b=0implies t'=frac{1 }{ alpha z^2+b}$$ which looks to be quite simple.
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Let $y=z-a+b t$ to make the equation
$$z'-alpha z^2-b=0$$ that is to say $$frac 1 {t'}-alpha z^2-b=0implies t'=frac{1 }{ alpha z^2+b}$$ which looks to be quite simple.
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Let $y=z-a+b t$ to make the equation
$$z'-alpha z^2-b=0$$ that is to say $$frac 1 {t'}-alpha z^2-b=0implies t'=frac{1 }{ alpha z^2+b}$$ which looks to be quite simple.
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Let $y=z-a+b t$ to make the equation
$$z'-alpha z^2-b=0$$ that is to say $$frac 1 {t'}-alpha z^2-b=0implies t'=frac{1 }{ alpha z^2+b}$$ which looks to be quite simple.
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 29 at 7:00
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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