solving hard differential equation

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3
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I have the following differential equation that I try to solve:



beginequation
(1 + t^2) doty = 2yt + t^2 + t^4
endequation



what I do is to first put all t's on one side:



beginequation
doty = frac2yt + t^2 + t^41+t^2
endequation



and then to integrate both sides



beginequation
int doty dy = int frac2yt + t^2 + t^41+t^2 dt
endequation



and well now I am stuck because I dont really know how to evaluate these complex integrals. I think I should use the method by parts but whenever I try I think I am making mistake somewhere and I am not sure if my steps up till now are correct as well.






differential-equations







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  • 1




    I think you should be able to get solution in terms of power series expansion relatively easily, but maybe that is not what you want.
    – mathreadler
    45 mins ago











  • @mathreadler and how would I do that? Well I am not sure how to solve it to be honest. I dont need to use any particular method I think
    – 1muflon1
    44 mins ago






  • 2




    You can't integrate the right side with that $y$ in there. This is a linear equation. Express it in the form $y'+p(t)y = g(t)$.
    – B. Goddard
    42 mins ago










  • You can search for some vector space where derivative and multiplicaiton with polynomial are linear operations. Then your equation becomes a linear equation system in this new space.
    – mathreadler
    42 mins ago










  • @B.Goddard So if I understand you correctly I should express it as: $y'- 2yt =fract^2+t^41-t^2$? Also how would this help? now the t is on the other side
    – 1muflon1
    39 mins ago















up vote
3
down vote

favorite












I have the following differential equation that I try to solve:



beginequation
(1 + t^2) doty = 2yt + t^2 + t^4
endequation



what I do is to first put all t's on one side:



beginequation
doty = frac2yt + t^2 + t^41+t^2
endequation



and then to integrate both sides



beginequation
int doty dy = int frac2yt + t^2 + t^41+t^2 dt
endequation



and well now I am stuck because I dont really know how to evaluate these complex integrals. I think I should use the method by parts but whenever I try I think I am making mistake somewhere and I am not sure if my steps up till now are correct as well.






differential-equations







share|cite|improve this question

















  • 1




    I think you should be able to get solution in terms of power series expansion relatively easily, but maybe that is not what you want.
    – mathreadler
    45 mins ago











  • @mathreadler and how would I do that? Well I am not sure how to solve it to be honest. I dont need to use any particular method I think
    – 1muflon1
    44 mins ago






  • 2




    You can't integrate the right side with that $y$ in there. This is a linear equation. Express it in the form $y'+p(t)y = g(t)$.
    – B. Goddard
    42 mins ago










  • You can search for some vector space where derivative and multiplicaiton with polynomial are linear operations. Then your equation becomes a linear equation system in this new space.
    – mathreadler
    42 mins ago










  • @B.Goddard So if I understand you correctly I should express it as: $y'- 2yt =fract^2+t^41-t^2$? Also how would this help? now the t is on the other side
    – 1muflon1
    39 mins ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have the following differential equation that I try to solve:



beginequation
(1 + t^2) doty = 2yt + t^2 + t^4
endequation



what I do is to first put all t's on one side:



beginequation
doty = frac2yt + t^2 + t^41+t^2
endequation



and then to integrate both sides



beginequation
int doty dy = int frac2yt + t^2 + t^41+t^2 dt
endequation



and well now I am stuck because I dont really know how to evaluate these complex integrals. I think I should use the method by parts but whenever I try I think I am making mistake somewhere and I am not sure if my steps up till now are correct as well.






differential-equations







share|cite|improve this question













I have the following differential equation that I try to solve:



beginequation
(1 + t^2) doty = 2yt + t^2 + t^4
endequation



what I do is to first put all t's on one side:



beginequation
doty = frac2yt + t^2 + t^41+t^2
endequation



and then to integrate both sides



beginequation
int doty dy = int frac2yt + t^2 + t^41+t^2 dt
endequation



and well now I am stuck because I dont really know how to evaluate these complex integrals. I think I should use the method by parts but whenever I try I think I am making mistake somewhere and I am not sure if my steps up till now are correct as well.






differential-equations




differential-equations






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 47 mins ago









1muflon1

967




967







  • 1




    I think you should be able to get solution in terms of power series expansion relatively easily, but maybe that is not what you want.
    – mathreadler
    45 mins ago











  • @mathreadler and how would I do that? Well I am not sure how to solve it to be honest. I dont need to use any particular method I think
    – 1muflon1
    44 mins ago






  • 2




    You can't integrate the right side with that $y$ in there. This is a linear equation. Express it in the form $y'+p(t)y = g(t)$.
    – B. Goddard
    42 mins ago










  • You can search for some vector space where derivative and multiplicaiton with polynomial are linear operations. Then your equation becomes a linear equation system in this new space.
    – mathreadler
    42 mins ago










  • @B.Goddard So if I understand you correctly I should express it as: $y'- 2yt =fract^2+t^41-t^2$? Also how would this help? now the t is on the other side
    – 1muflon1
    39 mins ago













  • 1




    I think you should be able to get solution in terms of power series expansion relatively easily, but maybe that is not what you want.
    – mathreadler
    45 mins ago











  • @mathreadler and how would I do that? Well I am not sure how to solve it to be honest. I dont need to use any particular method I think
    – 1muflon1
    44 mins ago






  • 2




    You can't integrate the right side with that $y$ in there. This is a linear equation. Express it in the form $y'+p(t)y = g(t)$.
    – B. Goddard
    42 mins ago










  • You can search for some vector space where derivative and multiplicaiton with polynomial are linear operations. Then your equation becomes a linear equation system in this new space.
    – mathreadler
    42 mins ago










  • @B.Goddard So if I understand you correctly I should express it as: $y'- 2yt =fract^2+t^41-t^2$? Also how would this help? now the t is on the other side
    – 1muflon1
    39 mins ago








1




1




I think you should be able to get solution in terms of power series expansion relatively easily, but maybe that is not what you want.
– mathreadler
45 mins ago





I think you should be able to get solution in terms of power series expansion relatively easily, but maybe that is not what you want.
– mathreadler
45 mins ago













@mathreadler and how would I do that? Well I am not sure how to solve it to be honest. I dont need to use any particular method I think
– 1muflon1
44 mins ago




@mathreadler and how would I do that? Well I am not sure how to solve it to be honest. I dont need to use any particular method I think
– 1muflon1
44 mins ago




2




2




You can't integrate the right side with that $y$ in there. This is a linear equation. Express it in the form $y'+p(t)y = g(t)$.
– B. Goddard
42 mins ago




You can't integrate the right side with that $y$ in there. This is a linear equation. Express it in the form $y'+p(t)y = g(t)$.
– B. Goddard
42 mins ago












You can search for some vector space where derivative and multiplicaiton with polynomial are linear operations. Then your equation becomes a linear equation system in this new space.
– mathreadler
42 mins ago




You can search for some vector space where derivative and multiplicaiton with polynomial are linear operations. Then your equation becomes a linear equation system in this new space.
– mathreadler
42 mins ago












@B.Goddard So if I understand you correctly I should express it as: $y'- 2yt =fract^2+t^41-t^2$? Also how would this help? now the t is on the other side
– 1muflon1
39 mins ago





@B.Goddard So if I understand you correctly I should express it as: $y'- 2yt =fract^2+t^41-t^2$? Also how would this help? now the t is on the other side
– 1muflon1
39 mins ago











2 Answers
2






active

oldest

votes

















up vote
3
down vote













The homogeneous equation is separable,



$$fracdot yy=frac2tt^2+1$$ and integrates as



$$log y=log(t^2+1)+C$$ or



$$y=C(t^2+1).$$



Now by variation of the constant, after simplification,



$$(t^2+1)^2C'(t)=t^2+t^4$$ then



$$C'(t)=fract^2t^2+1$$ and



$$C(t)=t-arctan t+D.$$



Finally,



$$y(t)=(t-arctan t+D)(t^2+1).$$






share|cite|improve this answer






















  • Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
    – 1muflon1
    21 mins ago










  • @1muflon1: the equation is not seaprable. Read my answer carefully.
    – Yves Daoust
    19 mins ago


















up vote
1
down vote













Rewrite the equation:



$$y'-frac2t1+t^2y = fract^4+t^2t^2+1 = t^2.$$



The standard technique for linear, first-order DE's is to multiply by the integrating factor



$$mu = expleft(int p(t) ; dtright) = expleft(int frac-2t1+t^2 ; dtright)$$



$$=exp(-ln(t^2+1)) = frac1t^2+1.$$



So now the DE is



$$frac1t^2+1y' + frac-2t1+t^2y = fract^21+t^2. $$



The left side is the derivative of $frac1t^2+1y$, so you have



$$fracyt^2+1 = int fract^2t^2+1 ; dt.$$






share|cite|improve this answer






















  • You mishandled a coefficient, making the solution go in a wrong direction.
    – Yves Daoust
    20 mins ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













The homogeneous equation is separable,



$$fracdot yy=frac2tt^2+1$$ and integrates as



$$log y=log(t^2+1)+C$$ or



$$y=C(t^2+1).$$



Now by variation of the constant, after simplification,



$$(t^2+1)^2C'(t)=t^2+t^4$$ then



$$C'(t)=fract^2t^2+1$$ and



$$C(t)=t-arctan t+D.$$



Finally,



$$y(t)=(t-arctan t+D)(t^2+1).$$






share|cite|improve this answer






















  • Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
    – 1muflon1
    21 mins ago










  • @1muflon1: the equation is not seaprable. Read my answer carefully.
    – Yves Daoust
    19 mins ago















up vote
3
down vote













The homogeneous equation is separable,



$$fracdot yy=frac2tt^2+1$$ and integrates as



$$log y=log(t^2+1)+C$$ or



$$y=C(t^2+1).$$



Now by variation of the constant, after simplification,



$$(t^2+1)^2C'(t)=t^2+t^4$$ then



$$C'(t)=fract^2t^2+1$$ and



$$C(t)=t-arctan t+D.$$



Finally,



$$y(t)=(t-arctan t+D)(t^2+1).$$






share|cite|improve this answer






















  • Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
    – 1muflon1
    21 mins ago










  • @1muflon1: the equation is not seaprable. Read my answer carefully.
    – Yves Daoust
    19 mins ago













up vote
3
down vote










up vote
3
down vote









The homogeneous equation is separable,



$$fracdot yy=frac2tt^2+1$$ and integrates as



$$log y=log(t^2+1)+C$$ or



$$y=C(t^2+1).$$



Now by variation of the constant, after simplification,



$$(t^2+1)^2C'(t)=t^2+t^4$$ then



$$C'(t)=fract^2t^2+1$$ and



$$C(t)=t-arctan t+D.$$



Finally,



$$y(t)=(t-arctan t+D)(t^2+1).$$






share|cite|improve this answer














The homogeneous equation is separable,



$$fracdot yy=frac2tt^2+1$$ and integrates as



$$log y=log(t^2+1)+C$$ or



$$y=C(t^2+1).$$



Now by variation of the constant, after simplification,



$$(t^2+1)^2C'(t)=t^2+t^4$$ then



$$C'(t)=fract^2t^2+1$$ and



$$C(t)=t-arctan t+D.$$



Finally,



$$y(t)=(t-arctan t+D)(t^2+1).$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 18 mins ago

























answered 36 mins ago









Yves Daoust

119k667215




119k667215











  • Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
    – 1muflon1
    21 mins ago










  • @1muflon1: the equation is not seaprable. Read my answer carefully.
    – Yves Daoust
    19 mins ago

















  • Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
    – 1muflon1
    21 mins ago










  • @1muflon1: the equation is not seaprable. Read my answer carefully.
    – Yves Daoust
    19 mins ago
















Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
– 1muflon1
21 mins ago




Hi thanks for the answer. How did you know that this equation is separable I thought that equation is separable only if I can get in a form $y(x) dy/dx = M(x)$ so only when all x in my case t can be separated and I can multiply the time derivative with the function
– 1muflon1
21 mins ago












@1muflon1: the equation is not seaprable. Read my answer carefully.
– Yves Daoust
19 mins ago





@1muflon1: the equation is not seaprable. Read my answer carefully.
– Yves Daoust
19 mins ago











up vote
1
down vote













Rewrite the equation:



$$y'-frac2t1+t^2y = fract^4+t^2t^2+1 = t^2.$$



The standard technique for linear, first-order DE's is to multiply by the integrating factor



$$mu = expleft(int p(t) ; dtright) = expleft(int frac-2t1+t^2 ; dtright)$$



$$=exp(-ln(t^2+1)) = frac1t^2+1.$$



So now the DE is



$$frac1t^2+1y' + frac-2t1+t^2y = fract^21+t^2. $$



The left side is the derivative of $frac1t^2+1y$, so you have



$$fracyt^2+1 = int fract^2t^2+1 ; dt.$$






share|cite|improve this answer






















  • You mishandled a coefficient, making the solution go in a wrong direction.
    – Yves Daoust
    20 mins ago














up vote
1
down vote













Rewrite the equation:



$$y'-frac2t1+t^2y = fract^4+t^2t^2+1 = t^2.$$



The standard technique for linear, first-order DE's is to multiply by the integrating factor



$$mu = expleft(int p(t) ; dtright) = expleft(int frac-2t1+t^2 ; dtright)$$



$$=exp(-ln(t^2+1)) = frac1t^2+1.$$



So now the DE is



$$frac1t^2+1y' + frac-2t1+t^2y = fract^21+t^2. $$



The left side is the derivative of $frac1t^2+1y$, so you have



$$fracyt^2+1 = int fract^2t^2+1 ; dt.$$






share|cite|improve this answer






















  • You mishandled a coefficient, making the solution go in a wrong direction.
    – Yves Daoust
    20 mins ago












up vote
1
down vote










up vote
1
down vote









Rewrite the equation:



$$y'-frac2t1+t^2y = fract^4+t^2t^2+1 = t^2.$$



The standard technique for linear, first-order DE's is to multiply by the integrating factor



$$mu = expleft(int p(t) ; dtright) = expleft(int frac-2t1+t^2 ; dtright)$$



$$=exp(-ln(t^2+1)) = frac1t^2+1.$$



So now the DE is



$$frac1t^2+1y' + frac-2t1+t^2y = fract^21+t^2. $$



The left side is the derivative of $frac1t^2+1y$, so you have



$$fracyt^2+1 = int fract^2t^2+1 ; dt.$$






share|cite|improve this answer














Rewrite the equation:



$$y'-frac2t1+t^2y = fract^4+t^2t^2+1 = t^2.$$



The standard technique for linear, first-order DE's is to multiply by the integrating factor



$$mu = expleft(int p(t) ; dtright) = expleft(int frac-2t1+t^2 ; dtright)$$



$$=exp(-ln(t^2+1)) = frac1t^2+1.$$



So now the DE is



$$frac1t^2+1y' + frac-2t1+t^2y = fract^21+t^2. $$



The left side is the derivative of $frac1t^2+1y$, so you have



$$fracyt^2+1 = int fract^2t^2+1 ; dt.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 mins ago

























answered 22 mins ago









B. Goddard

17.3k21338




17.3k21338











  • You mishandled a coefficient, making the solution go in a wrong direction.
    – Yves Daoust
    20 mins ago
















  • You mishandled a coefficient, making the solution go in a wrong direction.
    – Yves Daoust
    20 mins ago















You mishandled a coefficient, making the solution go in a wrong direction.
– Yves Daoust
20 mins ago




You mishandled a coefficient, making the solution go in a wrong direction.
– Yves Daoust
20 mins ago

















 

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