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How to solve this ODE $(y - x y^2) dx - (x + x^2 y) dy =0$?
2 ways to solve an ODE, different solutions…How can I solve this non separable ODE.How to solve this two variable Bernoulli equation ODE?How to solve this Ricatti-like ODEFirst order differential equationODE non separation questionHow do I solve this ODE question?Different ways to solve ODEHow to solve $y''(x) - frac2x^2 y(x) = 0$ by separation of variables?How to solve linear ODE $y'-axy=b$
$begingroup$
i have to solve this ODE: $(y - x y^2) dx - (x + x^2 y) dy =0$
I have tried by many methods (integrating factor,separation of variables, etc.. ) but i cant solve it.
ordinary-differential-equations
asked Mar 11 at 11:02
CatacrokerCatacroker
263
$endgroup$
add a comment |
$begingroup$
i have to solve this ODE: $(y - x y^2) dx - (x + x^2 y) dy =0$
I have tried by many methods (integrating factor,separation of variables, etc.. ) but i cant solve it.
ordinary-differential-equations
asked Mar 11 at 11:02
CatacrokerCatacroker
263
$endgroup$
add a comment |
$begingroup$
i have to solve this ODE: $(y - x y^2) dx - (x + x^2 y) dy =0$
I have tried by many methods (integrating factor,separation of variables, etc.. ) but i cant solve it.
ordinary-differential-equations
asked Mar 11 at 11:02
CatacrokerCatacroker
263
$endgroup$
i have to solve this ODE: $(y - x y^2) dx - (x + x^2 y) dy =0$
I have tried by many methods (integrating factor,separation of variables, etc.. ) but i cant solve it.
ordinary-differential-equations
ordinary-differential-equations
asked Mar 11 at 11:02
CatacrokerCatacroker
263
asked Mar 11 at 11:02
CatacrokerCatacroker
263
asked Mar 11 at 11:02
CatacrokerCatacroker
263
asked Mar 11 at 11:02
CatacrokerCatacroker
263
asked Mar 11 at 11:02
CatacrokerCatacroker
263
263
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint
Consider $$(y - x y^2) - (x + x^2 y), y' =0$$
and make $y=frac z x$ . This should give
$$frac2 zx-(z+1) z'=0$$ which does not look too bad.
Just for your curiosity, there is even an analytical solution $z(x)$ which uses a special function.
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Integrating factor is
$$mu=frac1xy.$$
Solution:
$$logleft( fracxyright) -x y=C.$$
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
$endgroup$
add a comment |
$begingroup$
$$(y - x y^2) dx - (x + x^2 y) dy =0to\ydx-xdy-xy(ydx+xdy)=0to \y^2d(xover y)=xyd(xy)$$by defining $v=xover y$ and $u=xy$ we obtain:$$uover vdv=uduto \dvover v=duto \ln v=u+C_1to\ v=Ce^uquad,quad C>0to \x=Cye^x yquad,quad C>0$$
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
Hint
Consider $$(y - x y^2) - (x + x^2 y), y' =0$$
and make $y=frac z x$ . This should give
$$frac2 zx-(z+1) z'=0$$ which does not look too bad.
Just for your curiosity, there is even an analytical solution $z(x)$ which uses a special function.
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Hint
Consider $$(y - x y^2) - (x + x^2 y), y' =0$$
and make $y=frac z x$ . This should give
$$frac2 zx-(z+1) z'=0$$ which does not look too bad.
Just for your curiosity, there is even an analytical solution $z(x)$ which uses a special function.
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Hint
Consider $$(y - x y^2) - (x + x^2 y), y' =0$$
and make $y=frac z x$ . This should give
$$frac2 zx-(z+1) z'=0$$ which does not look too bad.
Just for your curiosity, there is even an analytical solution $z(x)$ which uses a special function.
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
Hint
Consider $$(y - x y^2) - (x + x^2 y), y' =0$$
and make $y=frac z x$ . This should give
$$frac2 zx-(z+1) z'=0$$ which does not look too bad.
Just for your curiosity, there is even an analytical solution $z(x)$ which uses a special function.
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
answered Mar 11 at 11:18
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
add a comment |
add a comment |
$begingroup$
Integrating factor is
$$mu=frac1xy.$$
Solution:
$$logleft( fracxyright) -x y=C.$$
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
$endgroup$
add a comment |
$begingroup$
Integrating factor is
$$mu=frac1xy.$$
Solution:
$$logleft( fracxyright) -x y=C.$$
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
$endgroup$
add a comment |
$begingroup$
Integrating factor is
$$mu=frac1xy.$$
Solution:
$$logleft( fracxyright) -x y=C.$$
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
$endgroup$
Integrating factor is
$$mu=frac1xy.$$
Solution:
$$logleft( fracxyright) -x y=C.$$
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
answered Mar 11 at 11:44
Aleksas DomarkasAleksas Domarkas
1,47016
1,47016
add a comment |
add a comment |
$begingroup$
$$(y - x y^2) dx - (x + x^2 y) dy =0to\ydx-xdy-xy(ydx+xdy)=0to \y^2d(xover y)=xyd(xy)$$by defining $v=xover y$ and $u=xy$ we obtain:$$uover vdv=uduto \dvover v=duto \ln v=u+C_1to\ v=Ce^uquad,quad C>0to \x=Cye^x yquad,quad C>0$$
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
$endgroup$
add a comment |
$begingroup$
$$(y - x y^2) dx - (x + x^2 y) dy =0to\ydx-xdy-xy(ydx+xdy)=0to \y^2d(xover y)=xyd(xy)$$by defining $v=xover y$ and $u=xy$ we obtain:$$uover vdv=uduto \dvover v=duto \ln v=u+C_1to\ v=Ce^uquad,quad C>0to \x=Cye^x yquad,quad C>0$$
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
$endgroup$
add a comment |
$begingroup$
$$(y - x y^2) dx - (x + x^2 y) dy =0to\ydx-xdy-xy(ydx+xdy)=0to \y^2d(xover y)=xyd(xy)$$by defining $v=xover y$ and $u=xy$ we obtain:$$uover vdv=uduto \dvover v=duto \ln v=u+C_1to\ v=Ce^uquad,quad C>0to \x=Cye^x yquad,quad C>0$$
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
$endgroup$
$$(y - x y^2) dx - (x + x^2 y) dy =0to\ydx-xdy-xy(ydx+xdy)=0to \y^2d(xover y)=xyd(xy)$$by defining $v=xover y$ and $u=xy$ we obtain:$$uover vdv=uduto \dvover v=duto \ln v=u+C_1to\ v=Ce^uquad,quad C>0to \x=Cye^x yquad,quad C>0$$
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
answered Mar 11 at 11:59
Mostafa AyazMostafa Ayaz
15.8k3939
15.8k3939
add a comment |
add a comment |
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