Separating $y$'s and $x$'s
I am doing this initial value problem where I have the equation $y' + frac{3}{x} y=frac{cos(x)}{x^3} $. I know how to do these kinds of problems but I am having trouble getting the $x$ to the right side and the $y$ into the left side, any help would be appreciated.
differential-equations initial-value-problems
edited Dec 10 '18 at 3:46
platty
3,370320
add a comment |
I am doing this initial value problem where I have the equation $y' + frac{3}{x} y=frac{cos(x)}{x^3} $. I know how to do these kinds of problems but I am having trouble getting the $x$ to the right side and the $y$ into the left side, any help would be appreciated.
differential-equations initial-value-problems
edited Dec 10 '18 at 3:46
platty
3,370320
1
It is not respectful to the people who've written answers to delete your question.
– robjohn♦
Dec 10 '18 at 4:09
add a comment |
I am doing this initial value problem where I have the equation $y' + frac{3}{x} y=frac{cos(x)}{x^3} $. I know how to do these kinds of problems but I am having trouble getting the $x$ to the right side and the $y$ into the left side, any help would be appreciated.
differential-equations initial-value-problems
edited Dec 10 '18 at 3:46
platty
3,370320
I am doing this initial value problem where I have the equation $y' + frac{3}{x} y=frac{cos(x)}{x^3} $. I know how to do these kinds of problems but I am having trouble getting the $x$ to the right side and the $y$ into the left side, any help would be appreciated.
differential-equations initial-value-problems
differential-equations initial-value-problems
edited Dec 10 '18 at 3:46
platty
3,370320
edited Dec 10 '18 at 3:46
platty
3,370320
edited Dec 10 '18 at 3:46
platty
3,370320
edited Dec 10 '18 at 3:46
platty
3,370320
edited Dec 10 '18 at 3:46
platty
3,370320
3,370320
asked Dec 10 '18 at 2:15
user606496
1
It is not respectful to the people who've written answers to delete your question.
– robjohn♦
Dec 10 '18 at 4:09
add a comment |
1
It is not respectful to the people who've written answers to delete your question.
– robjohn♦
Dec 10 '18 at 4:09
1
1
It is not respectful to the people who've written answers to delete your question.
– robjohn♦
Dec 10 '18 at 4:09
It is not respectful to the people who've written answers to delete your question.
– robjohn♦
Dec 10 '18 at 4:09
add a comment |
1 Answer
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You'll have to treat this like a first order linear (non-separable) differential equation. Multiply both sides by $x^3$. You'll see that the left side looks like the result of a product rule expansion, giving you
$$[yx^3]’=cos(x)$$
Then, you can integrate, to yield
$$yx^3=-sin x + C$$
and divide by $x^3$ to get
$$y=-frac{sin x}{x^3} + Cx^{-3}$$
answered Dec 10 '18 at 2:26
D.R.
1,453620
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
|
show 1 more comment
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1 Answer
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You'll have to treat this like a first order linear (non-separable) differential equation. Multiply both sides by $x^3$. You'll see that the left side looks like the result of a product rule expansion, giving you
$$[yx^3]’=cos(x)$$
Then, you can integrate, to yield
$$yx^3=-sin x + C$$
and divide by $x^3$ to get
$$y=-frac{sin x}{x^3} + Cx^{-3}$$
answered Dec 10 '18 at 2:26
D.R.
1,453620
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
|
show 1 more comment
You'll have to treat this like a first order linear (non-separable) differential equation. Multiply both sides by $x^3$. You'll see that the left side looks like the result of a product rule expansion, giving you
$$[yx^3]’=cos(x)$$
Then, you can integrate, to yield
$$yx^3=-sin x + C$$
and divide by $x^3$ to get
$$y=-frac{sin x}{x^3} + Cx^{-3}$$
answered Dec 10 '18 at 2:26
D.R.
1,453620
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
|
show 1 more comment
You'll have to treat this like a first order linear (non-separable) differential equation. Multiply both sides by $x^3$. You'll see that the left side looks like the result of a product rule expansion, giving you
$$[yx^3]’=cos(x)$$
Then, you can integrate, to yield
$$yx^3=-sin x + C$$
and divide by $x^3$ to get
$$y=-frac{sin x}{x^3} + Cx^{-3}$$
answered Dec 10 '18 at 2:26
D.R.
1,453620
You'll have to treat this like a first order linear (non-separable) differential equation. Multiply both sides by $x^3$. You'll see that the left side looks like the result of a product rule expansion, giving you
$$[yx^3]’=cos(x)$$
Then, you can integrate, to yield
$$yx^3=-sin x + C$$
and divide by $x^3$ to get
$$y=-frac{sin x}{x^3} + Cx^{-3}$$
answered Dec 10 '18 at 2:26
D.R.
1,453620
edited Dec 10 '18 at 3:51
answered Dec 10 '18 at 2:26
D.R.
1,453620
answered Dec 10 '18 at 2:26
D.R.
1,453620
answered Dec 10 '18 at 2:26
D.R.
1,453620
1,453620
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
|
show 1 more comment
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
Hey thanks for answering. I just multiplied both by x^3, but what does it looking like the product rule expansion tell me? I would sitll have x and y in left side
– user606496
Dec 10 '18 at 3:33
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
@A.B. See if you can do it now
– D.R.
Dec 10 '18 at 3:41
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
No, I meant I did get what you have, but I don't see how I can remove the x^3?
– user606496
Dec 10 '18 at 3:49
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
Thank you, my last question is regarding what am I integrating in respect to. I am assuming the right side is in respect to dx, but what is left side in respect to?
– user606496
Dec 10 '18 at 3:54
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
@A.B. Also dx. y is a function of x, so the left side is just a function of x.
– D.R.
Dec 10 '18 at 3:56
|
show 1 more comment
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It is not respectful to the people who've written answers to delete your question.
– robjohn♦
Dec 10 '18 at 4:09