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How do I find the particular solution?

-1

$begingroup$

How do I find the particular solution for$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$for $-dfracpi2<x<dfracpi2$

given that $y(0)=9$.

I have now come to:

$ dfrac13dfracsin^3(x)cos(x)$$+c=9$

$ dfrac13dfracsin^3(x)cos(x)$ $=0$

Therefore $c=9$

ordinary-differential-equations

share|cite|improve this question

edited Mar 19 at 13:50

MathGeek101

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In the first order linear case you don't even need to split into homogeneous/particular, you can explicitly construct the solution to an IVP by using an integrating factor. In this case the integrating factor is really "already there" if you look at the equation carefully...
  $endgroup$
  – Ian
  Mar 19 at 13:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is a "non-answer", as $y'$ is still unknown.
  $endgroup$
  – Yves Daoust
  Mar 19 at 13:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't change the question silently. This all that has been written obsolete.
  $endgroup$
  – Yves Daoust
  Mar 19 at 16:38
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

How do I find the particular solution for$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$for $-dfracpi2<x<dfracpi2$

given that $y(0)=9$.

I have now come to:

$ dfrac13dfracsin^3(x)cos(x)$$+c=9$

$ dfrac13dfracsin^3(x)cos(x)$ $=0$

Therefore $c=9$

ordinary-differential-equations

share|cite|improve this question

edited Mar 19 at 13:50

MathGeek101

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In the first order linear case you don't even need to split into homogeneous/particular, you can explicitly construct the solution to an IVP by using an integrating factor. In this case the integrating factor is really "already there" if you look at the equation carefully...
  $endgroup$
  – Ian
  Mar 19 at 13:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer is a "non-answer", as $y'$ is still unknown.
  $endgroup$
  – Yves Daoust
  Mar 19 at 13:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't change the question silently. This all that has been written obsolete.
  $endgroup$
  – Yves Daoust
  Mar 19 at 16:38
  
  

  
  
  
  

add a comment | 

$begingroup$

How do I find the particular solution for$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$for $-dfracpi2<x<dfracpi2$

given that $y(0)=9$.

I have now come to:

$ dfrac13dfracsin^3(x)cos(x)$$+c=9$

$ dfrac13dfracsin^3(x)cos(x)$ $=0$

Therefore $c=9$

ordinary-differential-equations

share|cite|improve this question

edited Mar 19 at 13:50

MathGeek101

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

$endgroup$

share|cite|improve this question

edited Mar 19 at 13:50

MathGeek101

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

share|cite|improve this question

edited Mar 19 at 13:50

MathGeek101

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

asked Mar 19 at 12:58

MathGeek101MathGeek101

46

3 Answers
3

active

oldest

votes

2

$begingroup$

$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$
HINT :
$$d(cos(x) y)=sin^2(x)cos(x)dx$$
Then, integrate.

share|cite|improve this answer

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

$endgroup$

add a comment | 

2

$begingroup$

Read the equation as $$(ycos x)'=frac13(sin^3x)'$$

and integrate. This will even give you the general solution.

share|cite|improve this answer

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

1

$begingroup$

Firstly we need to find the Integrating Factor but before that divide the equation by $cos x$. I'll denote it by $mu$

$$beginalignedmu&=exp int dfrac-sin xcos xmathrm dx\&=exp ln cos x=cos xendaligned$$

Notice that this $mu$ is specifically chosen so that the ODE becomes: $$left(ycos xright)'=left(sin^2xcos xright)implies y=dfrac1cos xintsin^2xcos xmathrm dx$$

Now all you have to do is integrate the expressino and find the value of the constant that makes the initial condition $y(0)=9$ true. Can you proceed?

share|cite|improve this answer

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't have to delete your old answer. I was going to take off the downvote
  $endgroup$
  – Dylan
  Mar 19 at 13:40
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$
HINT :
$$d(cos(x) y)=sin^2(x)cos(x)dx$$
Then, integrate.

share|cite|improve this answer

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

$endgroup$

add a comment | 

2

$begingroup$

$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$
HINT :
$$d(cos(x) y)=sin^2(x)cos(x)dx$$
Then, integrate.

share|cite|improve this answer

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

$endgroup$

add a comment | 

$begingroup$

$$cos(x)fracmathrm dymathrm dx-sin(x)y=sin^2(x)cos(x)$$
HINT :
$$d(cos(x) y)=sin^2(x)cos(x)dx$$
Then, integrate.

share|cite|improve this answer

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

$endgroup$

share|cite|improve this answer

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

answered Mar 19 at 13:14

JJacquelinJJacquelin

45.3k21856

2

$begingroup$

Read the equation as $$(ycos x)'=frac13(sin^3x)'$$

and integrate. This will even give you the general solution.

share|cite|improve this answer

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

2

$begingroup$

Read the equation as $$(ycos x)'=frac13(sin^3x)'$$

and integrate. This will even give you the general solution.

share|cite|improve this answer

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

$begingroup$

Read the equation as $$(ycos x)'=frac13(sin^3x)'$$

and integrate. This will even give you the general solution.

share|cite|improve this answer

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

$endgroup$

share|cite|improve this answer

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

answered Mar 19 at 13:24

Yves DaoustYves Daoust

131k676229

1

$begingroup$

Firstly we need to find the Integrating Factor but before that divide the equation by $cos x$. I'll denote it by $mu$

$$beginalignedmu&=exp int dfrac-sin xcos xmathrm dx\&=exp ln cos x=cos xendaligned$$

Notice that this $mu$ is specifically chosen so that the ODE becomes: $$left(ycos xright)'=left(sin^2xcos xright)implies y=dfrac1cos xintsin^2xcos xmathrm dx$$

Now all you have to do is integrate the expressino and find the value of the constant that makes the initial condition $y(0)=9$ true. Can you proceed?

share|cite|improve this answer

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't have to delete your old answer. I was going to take off the downvote
  $endgroup$
  – Dylan
  Mar 19 at 13:40
  

  
  
  
  

add a comment | 

1

$begingroup$

Firstly we need to find the Integrating Factor but before that divide the equation by $cos x$. I'll denote it by $mu$

$$beginalignedmu&=exp int dfrac-sin xcos xmathrm dx\&=exp ln cos x=cos xendaligned$$

Notice that this $mu$ is specifically chosen so that the ODE becomes: $$left(ycos xright)'=left(sin^2xcos xright)implies y=dfrac1cos xintsin^2xcos xmathrm dx$$

Now all you have to do is integrate the expressino and find the value of the constant that makes the initial condition $y(0)=9$ true. Can you proceed?

share|cite|improve this answer

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't have to delete your old answer. I was going to take off the downvote
  $endgroup$
  – Dylan
  Mar 19 at 13:40
  

  
  
  
  

add a comment | 

$begingroup$

Firstly we need to find the Integrating Factor but before that divide the equation by $cos x$. I'll denote it by $mu$

$$beginalignedmu&=exp int dfrac-sin xcos xmathrm dx\&=exp ln cos x=cos xendaligned$$

Notice that this $mu$ is specifically chosen so that the ODE becomes: $$left(ycos xright)'=left(sin^2xcos xright)implies y=dfrac1cos xintsin^2xcos xmathrm dx$$

Now all you have to do is integrate the expressino and find the value of the constant that makes the initial condition $y(0)=9$ true. Can you proceed?

share|cite|improve this answer

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423

$endgroup$

share|cite|improve this answer

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423

answered Mar 19 at 13:35

Paras KhoslaParas Khosla

2,701423