A difficult differential equation $ y(2x^4+y)frac{dy}{dx} = (1-4xy^2)x^2$

up vote
2
down vote

favorite

How to solve the following differential equation?
$$ y(2x^4+y)dfrac{dy}{dx} = (1-4xy^2)x^2$$

No clue as to how to even begin. Hints?

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

add a comment |

up vote
2
down vote

favorite

How to solve the following differential equation?
$$ y(2x^4+y)dfrac{dy}{dx} = (1-4xy^2)x^2$$

No clue as to how to even begin. Hints?

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

add a comment |

up vote
2
down vote

favorite

How to solve the following differential equation?
$$ y(2x^4+y)dfrac{dy}{dx} = (1-4xy^2)x^2$$

No clue as to how to even begin. Hints?

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

How to solve the following differential equation?
$$ y(2x^4+y)dfrac{dy}{dx} = (1-4xy^2)x^2$$

No clue as to how to even begin. Hints?

calculus differential-equations

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

asked Sep 11 '13 at 4:14

Parth Thakkar

2,73521635

add a comment |

3 Answers
3

active

oldest

votes

up vote
5
down vote

accepted

Hint: it's an exact differential equation.

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

Please drop some more hint. This doesn't seem so straight forward to me.
– Parth Thakkar
Sep 11 '13 at 4:36

Well, I am familiar, but that's just it. I am only familiar. I haven't mastered it yet. I tried to bring it in an exact form, but I couldn't.
– Parth Thakkar
Sep 11 '13 at 5:09

1

Wait, I think there's some terminology issue here. By Exact method, do you mean writing the equation so that it is expressed as $ d( text{blob} ) = text{something integrable}$? I just googled the 'exact' method and what I got was something with partial derivatives. That, I do not know.
– Parth Thakkar
Sep 11 '13 at 5:12

1

You are writing $M dx + N dy = 0$: see tutorial.math.lamar.edu/Classes/DE/Exact.aspx for worked examples. Regards
– Amzoti
Sep 11 '13 at 5:19

Thanks a lot for that resource. Just began reading and seems nice! +1!
– Parth Thakkar
Sep 11 '13 at 5:27

|
show 1 more comment

up vote
3
down vote

$$y(2x^4+y)frac{dy}{dx}=(1-4xy^2)x^2$$

$$2x^4yfrac{dy}{dx}+y^2frac{dy}{dx}=x^2-4x^3 y^2$$

$$2x^4yfrac{dy}{dx}+4x^3y^2+y^2frac{dy}{dx}=x^2$$

$$frac{d}{dx}(x^4 y^2)+y^2frac{dy}{dx}=x^2$$

$$x^4y^2+frac{y^3}{3}=frac{x^3}{3}+C$$

edited Mar 5 at 21:33

mucciolo

1,9741819

answered Mar 5 at 21:16

Ibrahim

311

add a comment |

up vote
0
down vote

Hint:
Apply $M(x,y)=frac{du}{dx}$ and $ N(x,y)=frac{du}{dy}$

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

answered Sep 11 '13 at 12:39

Angelo

434

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
5
down vote

accepted

Hint: it's an exact differential equation.

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

Please drop some more hint. This doesn't seem so straight forward to me.
– Parth Thakkar
Sep 11 '13 at 4:36

Well, I am familiar, but that's just it. I am only familiar. I haven't mastered it yet. I tried to bring it in an exact form, but I couldn't.
– Parth Thakkar
Sep 11 '13 at 5:09

1

Wait, I think there's some terminology issue here. By Exact method, do you mean writing the equation so that it is expressed as $ d( text{blob} ) = text{something integrable}$? I just googled the 'exact' method and what I got was something with partial derivatives. That, I do not know.
– Parth Thakkar
Sep 11 '13 at 5:12

1

You are writing $M dx + N dy = 0$: see tutorial.math.lamar.edu/Classes/DE/Exact.aspx for worked examples. Regards
– Amzoti
Sep 11 '13 at 5:19

Thanks a lot for that resource. Just began reading and seems nice! +1!
– Parth Thakkar
Sep 11 '13 at 5:27

|
show 1 more comment

up vote
5
down vote

accepted

Hint: it's an exact differential equation.

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

Please drop some more hint. This doesn't seem so straight forward to me.
– Parth Thakkar
Sep 11 '13 at 4:36

Well, I am familiar, but that's just it. I am only familiar. I haven't mastered it yet. I tried to bring it in an exact form, but I couldn't.
– Parth Thakkar
Sep 11 '13 at 5:09

1

Wait, I think there's some terminology issue here. By Exact method, do you mean writing the equation so that it is expressed as $ d( text{blob} ) = text{something integrable}$? I just googled the 'exact' method and what I got was something with partial derivatives. That, I do not know.
– Parth Thakkar
Sep 11 '13 at 5:12

1

You are writing $M dx + N dy = 0$: see tutorial.math.lamar.edu/Classes/DE/Exact.aspx for worked examples. Regards
– Amzoti
Sep 11 '13 at 5:19

Thanks a lot for that resource. Just began reading and seems nice! +1!
– Parth Thakkar
Sep 11 '13 at 5:27

|
show 1 more comment

up vote
5
down vote

accepted

Hint: it's an exact differential equation.

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

Hint: it's an exact differential equation.

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

answered Sep 11 '13 at 4:16

Robert Israel

315k23206453

Please drop some more hint. This doesn't seem so straight forward to me.
– Parth Thakkar
Sep 11 '13 at 4:36

Well, I am familiar, but that's just it. I am only familiar. I haven't mastered it yet. I tried to bring it in an exact form, but I couldn't.
– Parth Thakkar
Sep 11 '13 at 5:09

1

Wait, I think there's some terminology issue here. By Exact method, do you mean writing the equation so that it is expressed as $ d( text{blob} ) = text{something integrable}$? I just googled the 'exact' method and what I got was something with partial derivatives. That, I do not know.
– Parth Thakkar
Sep 11 '13 at 5:12

1

You are writing $M dx + N dy = 0$: see tutorial.math.lamar.edu/Classes/DE/Exact.aspx for worked examples. Regards
– Amzoti
Sep 11 '13 at 5:19

Thanks a lot for that resource. Just began reading and seems nice! +1!
– Parth Thakkar
Sep 11 '13 at 5:27

|
show 1 more comment

Please drop some more hint. This doesn't seem so straight forward to me.
– Parth Thakkar
Sep 11 '13 at 4:36

Well, I am familiar, but that's just it. I am only familiar. I haven't mastered it yet. I tried to bring it in an exact form, but I couldn't.
– Parth Thakkar
Sep 11 '13 at 5:09

1

Wait, I think there's some terminology issue here. By Exact method, do you mean writing the equation so that it is expressed as $ d( text{blob} ) = text{something integrable}$? I just googled the 'exact' method and what I got was something with partial derivatives. That, I do not know.
– Parth Thakkar
Sep 11 '13 at 5:12

1

You are writing $M dx + N dy = 0$: see tutorial.math.lamar.edu/Classes/DE/Exact.aspx for worked examples. Regards
– Amzoti
Sep 11 '13 at 5:19

Thanks a lot for that resource. Just began reading and seems nice! +1!
– Parth Thakkar
Sep 11 '13 at 5:27

Please drop some more hint. This doesn't seem so straight forward to me.
– Parth Thakkar
Sep 11 '13 at 4:36

Well, I am familiar, but that's just it. I am only familiar. I haven't mastered it yet. I tried to bring it in an exact form, but I couldn't.
– Parth Thakkar
Sep 11 '13 at 5:09

Wait, I think there's some terminology issue here. By Exact method, do you mean writing the equation so that it is expressed as $ d( text{blob} ) = text{something integrable}$? I just googled the 'exact' method and what I got was something with partial derivatives. That, I do not know.
– Parth Thakkar
Sep 11 '13 at 5:12

You are writing $M dx + N dy = 0$: see tutorial.math.lamar.edu/Classes/DE/Exact.aspx for worked examples. Regards
– Amzoti
Sep 11 '13 at 5:19

Thanks a lot for that resource. Just began reading and seems nice! +1!
– Parth Thakkar
Sep 11 '13 at 5:27

|
show 1 more comment

up vote
3
down vote

$$y(2x^4+y)frac{dy}{dx}=(1-4xy^2)x^2$$

$$2x^4yfrac{dy}{dx}+y^2frac{dy}{dx}=x^2-4x^3 y^2$$

$$2x^4yfrac{dy}{dx}+4x^3y^2+y^2frac{dy}{dx}=x^2$$

$$frac{d}{dx}(x^4 y^2)+y^2frac{dy}{dx}=x^2$$

$$x^4y^2+frac{y^3}{3}=frac{x^3}{3}+C$$

edited Mar 5 at 21:33

mucciolo

1,9741819

answered Mar 5 at 21:16

Ibrahim

311

add a comment |

up vote
3
down vote

$$y(2x^4+y)frac{dy}{dx}=(1-4xy^2)x^2$$

$$2x^4yfrac{dy}{dx}+y^2frac{dy}{dx}=x^2-4x^3 y^2$$

$$2x^4yfrac{dy}{dx}+4x^3y^2+y^2frac{dy}{dx}=x^2$$

$$frac{d}{dx}(x^4 y^2)+y^2frac{dy}{dx}=x^2$$

$$x^4y^2+frac{y^3}{3}=frac{x^3}{3}+C$$

edited Mar 5 at 21:33

mucciolo

1,9741819

answered Mar 5 at 21:16

Ibrahim

311

add a comment |

up vote
3
down vote

$$y(2x^4+y)frac{dy}{dx}=(1-4xy^2)x^2$$

$$2x^4yfrac{dy}{dx}+y^2frac{dy}{dx}=x^2-4x^3 y^2$$

$$2x^4yfrac{dy}{dx}+4x^3y^2+y^2frac{dy}{dx}=x^2$$

$$frac{d}{dx}(x^4 y^2)+y^2frac{dy}{dx}=x^2$$

$$x^4y^2+frac{y^3}{3}=frac{x^3}{3}+C$$

edited Mar 5 at 21:33

mucciolo

1,9741819

answered Mar 5 at 21:16

Ibrahim

311

$$y(2x^4+y)frac{dy}{dx}=(1-4xy^2)x^2$$

$$2x^4yfrac{dy}{dx}+y^2frac{dy}{dx}=x^2-4x^3 y^2$$

$$2x^4yfrac{dy}{dx}+4x^3y^2+y^2frac{dy}{dx}=x^2$$

$$frac{d}{dx}(x^4 y^2)+y^2frac{dy}{dx}=x^2$$

$$x^4y^2+frac{y^3}{3}=frac{x^3}{3}+C$$

edited Mar 5 at 21:33

mucciolo

1,9741819

answered Mar 5 at 21:16

Ibrahim

311

edited Mar 5 at 21:33

mucciolo

1,9741819

edited Mar 5 at 21:33

mucciolo

1,9741819

edited Mar 5 at 21:33

mucciolo

1,9741819

answered Mar 5 at 21:16

Ibrahim

311

answered Mar 5 at 21:16

Ibrahim

311

answered Mar 5 at 21:16

Ibrahim

311

add a comment |

up vote
0
down vote

Hint:
Apply $M(x,y)=frac{du}{dx}$ and $ N(x,y)=frac{du}{dy}$

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

answered Sep 11 '13 at 12:39

Angelo

434

add a comment |

up vote
0
down vote

Hint:
Apply $M(x,y)=frac{du}{dx}$ and $ N(x,y)=frac{du}{dy}$

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

answered Sep 11 '13 at 12:39

Angelo

434

add a comment |

up vote
0
down vote

Hint:
Apply $M(x,y)=frac{du}{dx}$ and $ N(x,y)=frac{du}{dy}$

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

answered Sep 11 '13 at 12:39

Angelo

434

Hint:
Apply $M(x,y)=frac{du}{dx}$ and $ N(x,y)=frac{du}{dy}$

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

answered Sep 11 '13 at 12:39

Angelo

434

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

edited Mar 18 '17 at 10:42

SaudiBombsYemen

95110

answered Sep 11 '13 at 12:39

Angelo

434

answered Sep 11 '13 at 12:39

Angelo

434

answered Sep 11 '13 at 12:39

Angelo

434

add a comment |

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