Solve this differential equation dy/dx [closed]












-2












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Solve this differential equation where






ordinary-differential-equations differential







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closed as off-topic by Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo Feb 2 at 7:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Is that a demand? If so, then no; you solve it.
    $endgroup$
    – Mark Viola
    Feb 2 at 3:51












  • $begingroup$
    This is quite straightforward to solve. Show us you've put some effort into it and we'll help you where you got stuck.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 3:57






  • 1




    $begingroup$
    Your question is headed for closure, since (1) you didn't write out the equation in $LaTeX$, which is inconvenient for your readers; (2) you didn't provide any context--information about why this is a problem of interest, or what you have tried. Not to worry, it takes a while to get the hang of the culture around here. Anyway, I've tried to give you a pointer in my answer below. Welcome to MSE, and Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:00












  • $begingroup$
    I also added the "differential-equations" tag to your post.
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:23
















-2












$begingroup$


Solve this differential equation where






ordinary-differential-equations differential







share|cite|improve this question











$endgroup$



closed as off-topic by Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo Feb 2 at 7:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Is that a demand? If so, then no; you solve it.
    $endgroup$
    – Mark Viola
    Feb 2 at 3:51












  • $begingroup$
    This is quite straightforward to solve. Show us you've put some effort into it and we'll help you where you got stuck.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 3:57






  • 1




    $begingroup$
    Your question is headed for closure, since (1) you didn't write out the equation in $LaTeX$, which is inconvenient for your readers; (2) you didn't provide any context--information about why this is a problem of interest, or what you have tried. Not to worry, it takes a while to get the hang of the culture around here. Anyway, I've tried to give you a pointer in my answer below. Welcome to MSE, and Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:00












  • $begingroup$
    I also added the "differential-equations" tag to your post.
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:23














-2












-2








-2


1



$begingroup$


Solve this differential equation where






ordinary-differential-equations differential







share|cite|improve this question











$endgroup$




Solve this differential equation where






ordinary-differential-equations differential




ordinary-differential-equations differential






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 4:22









Robert Lewis

48.9k23168




48.9k23168










asked Feb 2 at 3:42









Udit Anjali SwaroopaUdit Anjali Swaroopa

41




41




closed as off-topic by Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo Feb 2 at 7:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo Feb 2 at 7:55


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Claude Leibovici, Mark Viola, Decaf-Math, Dave, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Is that a demand? If so, then no; you solve it.
    $endgroup$
    – Mark Viola
    Feb 2 at 3:51












  • $begingroup$
    This is quite straightforward to solve. Show us you've put some effort into it and we'll help you where you got stuck.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 3:57






  • 1




    $begingroup$
    Your question is headed for closure, since (1) you didn't write out the equation in $LaTeX$, which is inconvenient for your readers; (2) you didn't provide any context--information about why this is a problem of interest, or what you have tried. Not to worry, it takes a while to get the hang of the culture around here. Anyway, I've tried to give you a pointer in my answer below. Welcome to MSE, and Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:00












  • $begingroup$
    I also added the "differential-equations" tag to your post.
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:23


















  • $begingroup$
    Is that a demand? If so, then no; you solve it.
    $endgroup$
    – Mark Viola
    Feb 2 at 3:51












  • $begingroup$
    This is quite straightforward to solve. Show us you've put some effort into it and we'll help you where you got stuck.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 3:57






  • 1




    $begingroup$
    Your question is headed for closure, since (1) you didn't write out the equation in $LaTeX$, which is inconvenient for your readers; (2) you didn't provide any context--information about why this is a problem of interest, or what you have tried. Not to worry, it takes a while to get the hang of the culture around here. Anyway, I've tried to give you a pointer in my answer below. Welcome to MSE, and Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:00












  • $begingroup$
    I also added the "differential-equations" tag to your post.
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:23
















$begingroup$
Is that a demand? If so, then no; you solve it.
$endgroup$
– Mark Viola
Feb 2 at 3:51






$begingroup$
Is that a demand? If so, then no; you solve it.
$endgroup$
– Mark Viola
Feb 2 at 3:51














$begingroup$
This is quite straightforward to solve. Show us you've put some effort into it and we'll help you where you got stuck.
$endgroup$
– Rhys Hughes
Feb 2 at 3:57




$begingroup$
This is quite straightforward to solve. Show us you've put some effort into it and we'll help you where you got stuck.
$endgroup$
– Rhys Hughes
Feb 2 at 3:57




1




1




$begingroup$
Your question is headed for closure, since (1) you didn't write out the equation in $LaTeX$, which is inconvenient for your readers; (2) you didn't provide any context--information about why this is a problem of interest, or what you have tried. Not to worry, it takes a while to get the hang of the culture around here. Anyway, I've tried to give you a pointer in my answer below. Welcome to MSE, and Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 4:00






$begingroup$
Your question is headed for closure, since (1) you didn't write out the equation in $LaTeX$, which is inconvenient for your readers; (2) you didn't provide any context--information about why this is a problem of interest, or what you have tried. Not to worry, it takes a while to get the hang of the culture around here. Anyway, I've tried to give you a pointer in my answer below. Welcome to MSE, and Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 4:00














$begingroup$
I also added the "differential-equations" tag to your post.
$endgroup$
– Robert Lewis
Feb 2 at 4:23




$begingroup$
I also added the "differential-equations" tag to your post.
$endgroup$
– Robert Lewis
Feb 2 at 4:23










1 Answer
1






active

oldest

votes


















3












$begingroup$

The equation may be transformed to a linear one as follows:



From



$dfrac{dy}{dx} + ytan x = y^3 cos x, tag 1$



write, after a multiplication by $y^{-3}$,



$y^{-3}dfrac{dy}{dx} + y^{-2}tan x = cos x, tag 2$



set



$z = y^{-2}; tag 3$



then



$dfrac{dz}{dx} = -2y^{-3} dfrac{dy}{dx}; tag 4$



(2) may thus be written



$-dfrac{1}{2} dfrac{dz}{dx} + z tan x = cos x, tag 5$



that is,



$dfrac{dz}{dx} -2(tan x)z = -2cos x, tag 6$



a linear equation with a known solution, which I leave to my readers to discover; it's all over the literature, and all over this website.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:01












  • $begingroup$
    @RhysHughes: which side?
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:02






  • 1




    $begingroup$
    Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:04


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The equation may be transformed to a linear one as follows:



From



$dfrac{dy}{dx} + ytan x = y^3 cos x, tag 1$



write, after a multiplication by $y^{-3}$,



$y^{-3}dfrac{dy}{dx} + y^{-2}tan x = cos x, tag 2$



set



$z = y^{-2}; tag 3$



then



$dfrac{dz}{dx} = -2y^{-3} dfrac{dy}{dx}; tag 4$



(2) may thus be written



$-dfrac{1}{2} dfrac{dz}{dx} + z tan x = cos x, tag 5$



that is,



$dfrac{dz}{dx} -2(tan x)z = -2cos x, tag 6$



a linear equation with a known solution, which I leave to my readers to discover; it's all over the literature, and all over this website.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:01












  • $begingroup$
    @RhysHughes: which side?
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:02






  • 1




    $begingroup$
    Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:04
















3












$begingroup$

The equation may be transformed to a linear one as follows:



From



$dfrac{dy}{dx} + ytan x = y^3 cos x, tag 1$



write, after a multiplication by $y^{-3}$,



$y^{-3}dfrac{dy}{dx} + y^{-2}tan x = cos x, tag 2$



set



$z = y^{-2}; tag 3$



then



$dfrac{dz}{dx} = -2y^{-3} dfrac{dy}{dx}; tag 4$



(2) may thus be written



$-dfrac{1}{2} dfrac{dz}{dx} + z tan x = cos x, tag 5$



that is,



$dfrac{dz}{dx} -2(tan x)z = -2cos x, tag 6$



a linear equation with a known solution, which I leave to my readers to discover; it's all over the literature, and all over this website.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:01












  • $begingroup$
    @RhysHughes: which side?
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:02






  • 1




    $begingroup$
    Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:04














3












3








3





$begingroup$

The equation may be transformed to a linear one as follows:



From



$dfrac{dy}{dx} + ytan x = y^3 cos x, tag 1$



write, after a multiplication by $y^{-3}$,



$y^{-3}dfrac{dy}{dx} + y^{-2}tan x = cos x, tag 2$



set



$z = y^{-2}; tag 3$



then



$dfrac{dz}{dx} = -2y^{-3} dfrac{dy}{dx}; tag 4$



(2) may thus be written



$-dfrac{1}{2} dfrac{dz}{dx} + z tan x = cos x, tag 5$



that is,



$dfrac{dz}{dx} -2(tan x)z = -2cos x, tag 6$



a linear equation with a known solution, which I leave to my readers to discover; it's all over the literature, and all over this website.






share|cite|improve this answer









$endgroup$



The equation may be transformed to a linear one as follows:



From



$dfrac{dy}{dx} + ytan x = y^3 cos x, tag 1$



write, after a multiplication by $y^{-3}$,



$y^{-3}dfrac{dy}{dx} + y^{-2}tan x = cos x, tag 2$



set



$z = y^{-2}; tag 3$



then



$dfrac{dz}{dx} = -2y^{-3} dfrac{dy}{dx}; tag 4$



(2) may thus be written



$-dfrac{1}{2} dfrac{dz}{dx} + z tan x = cos x, tag 5$



that is,



$dfrac{dz}{dx} -2(tan x)z = -2cos x, tag 6$



a linear equation with a known solution, which I leave to my readers to discover; it's all over the literature, and all over this website.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 2 at 3:56









Robert LewisRobert Lewis

48.9k23168




48.9k23168












  • $begingroup$
    Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:01












  • $begingroup$
    @RhysHughes: which side?
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:02






  • 1




    $begingroup$
    Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:04


















  • $begingroup$
    Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:01












  • $begingroup$
    @RhysHughes: which side?
    $endgroup$
    – Robert Lewis
    Feb 2 at 4:02






  • 1




    $begingroup$
    Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
    $endgroup$
    – Rhys Hughes
    Feb 2 at 4:04
















$begingroup$
Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
$endgroup$
– Rhys Hughes
Feb 2 at 4:01






$begingroup$
Please don't answer blatant PSQs, it's unhealthy for the site. Instead, I'd suggest you prepare your answer in advance, even leave it here if need be and request the OP to put effort in, at which point after they do so you can upload your answer.
$endgroup$
– Rhys Hughes
Feb 2 at 4:01














$begingroup$
@RhysHughes: which side?
$endgroup$
– Robert Lewis
Feb 2 at 4:02




$begingroup$
@RhysHughes: which side?
$endgroup$
– Robert Lewis
Feb 2 at 4:02




1




1




$begingroup$
Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
$endgroup$
– Rhys Hughes
Feb 2 at 4:04




$begingroup$
Lol typo my bad. Typing on a tablet with autocorrect is clunky as usual.
$endgroup$
– Rhys Hughes
Feb 2 at 4:04



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