Nonlinear differential equations, method
$begingroup$
There is a nonlinear differential equation $$ F(x, y, y',...,y^{(n)} )=0$$ One method to solve it is to use substitution $y'=yz$, but before that, there is one condition that has to be satisfied and that is $F(x, t_y, t_y',...,t_y^{(n)} )=t^{n}F(x, y, y',...,y^{(n)} )$. I found this in some notebooks, and I simply don't get what this $t$ represents.
Is anyone familiar with this method? I couldn't find anything on the internet, so I would be very grateful if somebody could explain this equality to me.
To be clearer, there is an example:
$$y(xy''+y')=xy'^2(1-x)$$
Checking the condition: $t_y (xt_y ''+t_y ')=xt^2y'^2(1-x)$
Substitution: $y'=yz$.
ordinary-differential-equations
asked Jan 6 at 17:20
user25568user25568
1088
$endgroup$
add a comment |
$begingroup$
There is a nonlinear differential equation $$ F(x, y, y',...,y^{(n)} )=0$$ One method to solve it is to use substitution $y'=yz$, but before that, there is one condition that has to be satisfied and that is $F(x, t_y, t_y',...,t_y^{(n)} )=t^{n}F(x, y, y',...,y^{(n)} )$. I found this in some notebooks, and I simply don't get what this $t$ represents.
Is anyone familiar with this method? I couldn't find anything on the internet, so I would be very grateful if somebody could explain this equality to me.
To be clearer, there is an example:
$$y(xy''+y')=xy'^2(1-x)$$
Checking the condition: $t_y (xt_y ''+t_y ')=xt^2y'^2(1-x)$
Substitution: $y'=yz$.
ordinary-differential-equations
asked Jan 6 at 17:20
user25568user25568
1088
$endgroup$
add a comment |
$begingroup$
There is a nonlinear differential equation $$ F(x, y, y',...,y^{(n)} )=0$$ One method to solve it is to use substitution $y'=yz$, but before that, there is one condition that has to be satisfied and that is $F(x, t_y, t_y',...,t_y^{(n)} )=t^{n}F(x, y, y',...,y^{(n)} )$. I found this in some notebooks, and I simply don't get what this $t$ represents.
Is anyone familiar with this method? I couldn't find anything on the internet, so I would be very grateful if somebody could explain this equality to me.
To be clearer, there is an example:
$$y(xy''+y')=xy'^2(1-x)$$
Checking the condition: $t_y (xt_y ''+t_y ')=xt^2y'^2(1-x)$
Substitution: $y'=yz$.
ordinary-differential-equations
asked Jan 6 at 17:20
user25568user25568
1088
$endgroup$
There is a nonlinear differential equation $$ F(x, y, y',...,y^{(n)} )=0$$ One method to solve it is to use substitution $y'=yz$, but before that, there is one condition that has to be satisfied and that is $F(x, t_y, t_y',...,t_y^{(n)} )=t^{n}F(x, y, y',...,y^{(n)} )$. I found this in some notebooks, and I simply don't get what this $t$ represents.
Is anyone familiar with this method? I couldn't find anything on the internet, so I would be very grateful if somebody could explain this equality to me.
To be clearer, there is an example:
$$y(xy''+y')=xy'^2(1-x)$$
Checking the condition: $t_y (xt_y ''+t_y ')=xt^2y'^2(1-x)$
Substitution: $y'=yz$.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 6 at 17:20
user25568user25568
1088
asked Jan 6 at 17:20
user25568user25568
1088
asked Jan 6 at 17:20
user25568user25568
1088
asked Jan 6 at 17:20
user25568user25568
1088
asked Jan 6 at 17:20
user25568user25568
1088
1088
add a comment |
add a comment |
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