Solution of the equation $dy/dx = 3yx^2/(y^3+3y^4)$?Differential Equation: Modifying Particular SolutionHow to justify the choice of particular solution?General solution to a second order nonhomogeneous differential equation.Analytic solution to Poisson equationParticular solution to nonhomogeneneous 2nd order ODESolve the differential equation $fracdydx + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution ReviewHomogeneous Solution of a 4th Order Differential EquationFind the solution of the equation given below :Why do we get particular solution when solving non homogeneous differential equation and not general solutionDiff Eq: Find the general solution of $y'= y^2 -1$
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Solution of the equation $dy/dx = 3yx^2/(y^3+3y^4)$?
Differential Equation: Modifying Particular SolutionHow to justify the choice of particular solution?General solution to a second order nonhomogeneous differential equation.Analytic solution to Poisson equationParticular solution to nonhomogeneneous 2nd order ODESolve the differential equation $fracdydx + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution ReviewHomogeneous Solution of a 4th Order Differential EquationFind the solution of the equation given below :Why do we get particular solution when solving non homogeneous differential equation and not general solutionDiff Eq: Find the general solution of $y'= y^2 -1$
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How do I find solution of the equation $dy/dx = 3yx^2/(y^3+3y^4)$? The variables can't be separated, the equation is not homogeneous and the equation can't be put in $dy/dx + Py = Q$ form. How then do I solve it?
calculus ordinary-differential-equations
asked Mar 29 at 1:21
HemaHema
6621213
$endgroup$
add a comment |
$begingroup$
How do I find solution of the equation $dy/dx = 3yx^2/(y^3+3y^4)$? The variables can't be separated, the equation is not homogeneous and the equation can't be put in $dy/dx + Py = Q$ form. How then do I solve it?
calculus ordinary-differential-equations
asked Mar 29 at 1:21
HemaHema
6621213
$endgroup$
1
$begingroup$
Why have you not cancelled the factor of $y$ in the second term ? ... Are you missing some brackets begineqnarray* fracdydx= frac3x^2ycolorred( y^3+3y^4 colorred) endeqnarray* would be much easier.
$endgroup$
– Donald Splutterwit
Mar 29 at 1:32
2
$begingroup$
Why not begin by simplifying it to $dy/dx = 3 (x^2 / y^2) + 3 y^4$? You can then find an integrating factor.
$endgroup$
– avs
Mar 29 at 1:33
$begingroup$
@avs in our syllabus the only IF taught has been for dy/dx + Py = Q form
$endgroup$
– Hema
Mar 29 at 2:00
4
$begingroup$
This is a separable equation. $(y^2 + 3 y^3); dy = 3 x^2 ; dx$.
$endgroup$
– Robert Israel
Mar 29 at 2:12
2
$begingroup$
@Hema, if the parenthesization suggested by Donald Splutterwit is correct, then (1) I think you need to practice algebra so as to get your use of parentheses correct (skipping the parentheses marked in red produces a completely different ODE), and (2) the equation can be solved by separation of variables.
$endgroup$
– avs
Mar 29 at 3:10
add a comment |
$begingroup$
How do I find solution of the equation $dy/dx = 3yx^2/(y^3+3y^4)$? The variables can't be separated, the equation is not homogeneous and the equation can't be put in $dy/dx + Py = Q$ form. How then do I solve it?
calculus ordinary-differential-equations
asked Mar 29 at 1:21
HemaHema
6621213
$endgroup$
How do I find solution of the equation $dy/dx = 3yx^2/(y^3+3y^4)$? The variables can't be separated, the equation is not homogeneous and the equation can't be put in $dy/dx + Py = Q$ form. How then do I solve it?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Mar 29 at 1:21
HemaHema
6621213
asked Mar 29 at 1:21
HemaHema
6621213
edited Mar 29 at 1:59
Hema
asked Mar 29 at 1:21
HemaHema
6621213
asked Mar 29 at 1:21
HemaHema
6621213
asked Mar 29 at 1:21
HemaHema
6621213
6621213
1
$begingroup$
Why have you not cancelled the factor of $y$ in the second term ? ... Are you missing some brackets begineqnarray* fracdydx= frac3x^2ycolorred( y^3+3y^4 colorred) endeqnarray* would be much easier.
$endgroup$
– Donald Splutterwit
Mar 29 at 1:32
2
$begingroup$
Why not begin by simplifying it to $dy/dx = 3 (x^2 / y^2) + 3 y^4$? You can then find an integrating factor.
$endgroup$
– avs
Mar 29 at 1:33
$begingroup$
@avs in our syllabus the only IF taught has been for dy/dx + Py = Q form
$endgroup$
– Hema
Mar 29 at 2:00
4
$begingroup$
This is a separable equation. $(y^2 + 3 y^3); dy = 3 x^2 ; dx$.
$endgroup$
– Robert Israel
Mar 29 at 2:12
2
$begingroup$
@Hema, if the parenthesization suggested by Donald Splutterwit is correct, then (1) I think you need to practice algebra so as to get your use of parentheses correct (skipping the parentheses marked in red produces a completely different ODE), and (2) the equation can be solved by separation of variables.
$endgroup$
– avs
Mar 29 at 3:10
add a comment |
1
$begingroup$
Why have you not cancelled the factor of $y$ in the second term ? ... Are you missing some brackets begineqnarray* fracdydx= frac3x^2ycolorred( y^3+3y^4 colorred) endeqnarray* would be much easier.
$endgroup$
– Donald Splutterwit
Mar 29 at 1:32
2
$begingroup$
Why not begin by simplifying it to $dy/dx = 3 (x^2 / y^2) + 3 y^4$? You can then find an integrating factor.
$endgroup$
– avs
Mar 29 at 1:33
$begingroup$
@avs in our syllabus the only IF taught has been for dy/dx + Py = Q form
$endgroup$
– Hema
Mar 29 at 2:00
4
$begingroup$
This is a separable equation. $(y^2 + 3 y^3); dy = 3 x^2 ; dx$.
$endgroup$
– Robert Israel
Mar 29 at 2:12
2
$begingroup$
@Hema, if the parenthesization suggested by Donald Splutterwit is correct, then (1) I think you need to practice algebra so as to get your use of parentheses correct (skipping the parentheses marked in red produces a completely different ODE), and (2) the equation can be solved by separation of variables.
$endgroup$
– avs
Mar 29 at 3:10
1
1
$begingroup$
Why have you not cancelled the factor of $y$ in the second term ? ... Are you missing some brackets begineqnarray* fracdydx= frac3x^2ycolorred( y^3+3y^4 colorred) endeqnarray* would be much easier.
$endgroup$
– Donald Splutterwit
Mar 29 at 1:32
$begingroup$
Why have you not cancelled the factor of $y$ in the second term ? ... Are you missing some brackets begineqnarray* fracdydx= frac3x^2ycolorred( y^3+3y^4 colorred) endeqnarray* would be much easier.
$endgroup$
– Donald Splutterwit
Mar 29 at 1:32
2
2
$begingroup$
Why not begin by simplifying it to $dy/dx = 3 (x^2 / y^2) + 3 y^4$? You can then find an integrating factor.
$endgroup$
– avs
Mar 29 at 1:33
$begingroup$
Why not begin by simplifying it to $dy/dx = 3 (x^2 / y^2) + 3 y^4$? You can then find an integrating factor.
$endgroup$
– avs
Mar 29 at 1:33
$begingroup$
@avs in our syllabus the only IF taught has been for dy/dx + Py = Q form
$endgroup$
– Hema
Mar 29 at 2:00
$begingroup$
@avs in our syllabus the only IF taught has been for dy/dx + Py = Q form
$endgroup$
– Hema
Mar 29 at 2:00
4
4
$begingroup$
This is a separable equation. $(y^2 + 3 y^3); dy = 3 x^2 ; dx$.
$endgroup$
– Robert Israel
Mar 29 at 2:12
$begingroup$
This is a separable equation. $(y^2 + 3 y^3); dy = 3 x^2 ; dx$.
$endgroup$
– Robert Israel
Mar 29 at 2:12
2
2
$begingroup$
@Hema, if the parenthesization suggested by Donald Splutterwit is correct, then (1) I think you need to practice algebra so as to get your use of parentheses correct (skipping the parentheses marked in red produces a completely different ODE), and (2) the equation can be solved by separation of variables.
$endgroup$
– avs
Mar 29 at 3:10
$begingroup$
@Hema, if the parenthesization suggested by Donald Splutterwit is correct, then (1) I think you need to practice algebra so as to get your use of parentheses correct (skipping the parentheses marked in red produces a completely different ODE), and (2) the equation can be solved by separation of variables.
$endgroup$
– avs
Mar 29 at 3:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
It is separable:
$$fracy^3+3y^43ydy=x^2dx$$
So for $y neq 0$, solve $left(frac13y^2+ y^3right) dy = x^2dx$
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
$endgroup$
add a comment |
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$begingroup$
Hint:
It is separable:
$$fracy^3+3y^43ydy=x^2dx$$
So for $y neq 0$, solve $left(frac13y^2+ y^3right) dy = x^2dx$
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
$endgroup$
add a comment |
$begingroup$
Hint:
It is separable:
$$fracy^3+3y^43ydy=x^2dx$$
So for $y neq 0$, solve $left(frac13y^2+ y^3right) dy = x^2dx$
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
$endgroup$
add a comment |
$begingroup$
Hint:
It is separable:
$$fracy^3+3y^43ydy=x^2dx$$
So for $y neq 0$, solve $left(frac13y^2+ y^3right) dy = x^2dx$
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
$endgroup$
Hint:
It is separable:
$$fracy^3+3y^43ydy=x^2dx$$
So for $y neq 0$, solve $left(frac13y^2+ y^3right) dy = x^2dx$
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
answered Mar 29 at 14:49
Fer BelloraFer Bellora
1355
1355
add a comment |
add a comment |
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$begingroup$
Why have you not cancelled the factor of $y$ in the second term ? ... Are you missing some brackets begineqnarray* fracdydx= frac3x^2ycolorred( y^3+3y^4 colorred) endeqnarray* would be much easier.
$endgroup$
– Donald Splutterwit
Mar 29 at 1:32
2
$begingroup$
Why not begin by simplifying it to $dy/dx = 3 (x^2 / y^2) + 3 y^4$? You can then find an integrating factor.
$endgroup$
– avs
Mar 29 at 1:33
$begingroup$
@avs in our syllabus the only IF taught has been for dy/dx + Py = Q form
$endgroup$
– Hema
Mar 29 at 2:00
4
$begingroup$
This is a separable equation. $(y^2 + 3 y^3); dy = 3 x^2 ; dx$.
$endgroup$
– Robert Israel
Mar 29 at 2:12
2
$begingroup$
@Hema, if the parenthesization suggested by Donald Splutterwit is correct, then (1) I think you need to practice algebra so as to get your use of parentheses correct (skipping the parentheses marked in red produces a completely different ODE), and (2) the equation can be solved by separation of variables.
$endgroup$
– avs
Mar 29 at 3:10