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Use an appropriate change of variables to solve the differential equation.
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Use an appropriate change of variables to solve the differential equation.
$$tfracdydt-y=sqrtt^2+y^2$$
My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.
Any help would be appreciated especially if you could help us with step by step.
Thanks!
--
UPDATE:
$$tfracdydt-y=sqrtt^2+y^2$$
$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t
$$fracdydt=sqrt1+u^2+u=f(u)$$
$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...
We ended up with $ln|sqrt1+u^2|=ln|x|+c$
differential-equations derivatives change-of-variable
asked 1 hour ago
Taljana D
180110
add a comment |Â
up vote
1
down vote
favorite
Use an appropriate change of variables to solve the differential equation.
$$tfracdydt-y=sqrtt^2+y^2$$
My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.
Any help would be appreciated especially if you could help us with step by step.
Thanks!
--
UPDATE:
$$tfracdydt-y=sqrtt^2+y^2$$
$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t
$$fracdydt=sqrt1+u^2+u=f(u)$$
$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...
We ended up with $ln|sqrt1+u^2|=ln|x|+c$
differential-equations derivatives change-of-variable
asked 1 hour ago
Taljana D
180110
As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Use an appropriate change of variables to solve the differential equation.
$$tfracdydt-y=sqrtt^2+y^2$$
My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.
Any help would be appreciated especially if you could help us with step by step.
Thanks!
--
UPDATE:
$$tfracdydt-y=sqrtt^2+y^2$$
$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t
$$fracdydt=sqrt1+u^2+u=f(u)$$
$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...
We ended up with $ln|sqrt1+u^2|=ln|x|+c$
differential-equations derivatives change-of-variable
asked 1 hour ago
Taljana D
180110
Use an appropriate change of variables to solve the differential equation.
$$tfracdydt-y=sqrtt^2+y^2$$
My friend and I are trying to figure out how to solve this equation. Our professor has given us several methods but we aren't sure which to use because none of the equations are similar to this one.
Any help would be appreciated especially if you could help us with step by step.
Thanks!
--
UPDATE:
$$tfracdydt-y=sqrtt^2+y^2$$
$$fracdydt-fracyt=sqrt1+fracy^2t$$ where u= y/t
$$fracdydt=sqrt1+u^2+u=f(u)$$
$$f(u)-u=sqrt1+u^2+u-u=sqrt1+u^2$$
...
We ended up with $ln|sqrt1+u^2|=ln|x|+c$
differential-equations derivatives change-of-variable
differential-equations derivatives change-of-variable
asked 1 hour ago
Taljana D
180110
asked 1 hour ago
Taljana D
180110
edited 54 mins ago
asked 1 hour ago
Taljana D
180110
asked 1 hour ago
Taljana D
180110
asked 1 hour ago
Taljana D
180110
180110
As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago
add a comment |Â
As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago
As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago
As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
Divide by $t$ to obtain,
beginequation
fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
endequation
then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.
answered 1 hour ago
Andrew Krause
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you check what we've done above please!
– Taljana D
53 mins ago
add a comment |Â
up vote
2
down vote
HINT
One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.
answered 1 hour ago
gt6989b
31.1k22349
Can you check what we've done above please.
– Taljana D
52 mins ago
add a comment |Â
up vote
2
down vote
Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
Can you proceed?
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
Can you check what we've done above please!
– Taljana D
52 mins ago
add a comment |Â
up vote
1
down vote
By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes
$$fracdudt=frac1tsqrt1+u^2$$
which can be solved by separation of variables. To be exact
$$beginalign
fracdudt&=frac1tsqrt1+u^2\
fracdusqrt1+u^2&=fracdtt\
intfracdusqrt1+u^2&=intfracdtt\
operatornamearsinh(u)&=log t +c\
u&=sinh(log t+c)
endalign$$
and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.
answered 1 hour ago
mrtaurho
1,190320
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Divide by $t$ to obtain,
beginequation
fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
endequation
then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.
answered 1 hour ago
Andrew Krause
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you check what we've done above please!
– Taljana D
53 mins ago
add a comment |Â
up vote
4
down vote
Divide by $t$ to obtain,
beginequation
fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
endequation
then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.
answered 1 hour ago
Andrew Krause
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you check what we've done above please!
– Taljana D
53 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Divide by $t$ to obtain,
beginequation
fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
endequation
then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.
answered 1 hour ago
Andrew Krause
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Divide by $t$ to obtain,
beginequation
fracdydt - fracyt = sqrt1+left ( fracyt right )^2,
endequation
then use a change of variables $u = fracyt$. The transformed equation should be integrable using standard methods.
answered 1 hour ago
Andrew Krause
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Andrew Krause
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Andrew Krause
1412
answered 1 hour ago
Andrew Krause
1412
1412
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andrew Krause is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you check what we've done above please!
– Taljana D
53 mins ago
add a comment |Â
Could you check what we've done above please!
– Taljana D
53 mins ago
Could you check what we've done above please!
– Taljana D
53 mins ago
Could you check what we've done above please!
– Taljana D
53 mins ago
add a comment |Â
up vote
2
down vote
HINT
One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.
answered 1 hour ago
gt6989b
31.1k22349
Can you check what we've done above please.
– Taljana D
52 mins ago
add a comment |Â
up vote
2
down vote
HINT
One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.
answered 1 hour ago
gt6989b
31.1k22349
Can you check what we've done above please.
– Taljana D
52 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT
One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.
answered 1 hour ago
gt6989b
31.1k22349
HINT
One easy one to try is $y=zt$ so $y'=z't+z$ and the ODE becomes $$z't = sqrt1+z^2$$ which is separable.
answered 1 hour ago
gt6989b
31.1k22349
answered 1 hour ago
gt6989b
31.1k22349
answered 1 hour ago
gt6989b
31.1k22349
answered 1 hour ago
gt6989b
31.1k22349
31.1k22349
Can you check what we've done above please.
– Taljana D
52 mins ago
add a comment |Â
Can you check what we've done above please.
– Taljana D
52 mins ago
Can you check what we've done above please.
– Taljana D
52 mins ago
Can you check what we've done above please.
– Taljana D
52 mins ago
add a comment |Â
up vote
2
down vote
Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
Can you proceed?
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
Can you check what we've done above please!
– Taljana D
52 mins ago
add a comment |Â
up vote
2
down vote
Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
Can you proceed?
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
Can you check what we've done above please!
– Taljana D
52 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
Can you proceed?
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
Hint: Substitute $$y(t)=tv(t)$$ then you will get
$$tleft(tfracdv(t)dt+v(t)right)-tv(t)=sqrtt^2+t^2v(t)^2$$
Can you proceed?
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
answered 1 hour ago
Dr. Sonnhard Graubner
69.6k32761
69.6k32761
Can you check what we've done above please!
– Taljana D
52 mins ago
add a comment |Â
Can you check what we've done above please!
– Taljana D
52 mins ago
Can you check what we've done above please!
– Taljana D
52 mins ago
Can you check what we've done above please!
– Taljana D
52 mins ago
add a comment |Â
up vote
1
down vote
By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes
$$fracdudt=frac1tsqrt1+u^2$$
which can be solved by separation of variables. To be exact
$$beginalign
fracdudt&=frac1tsqrt1+u^2\
fracdusqrt1+u^2&=fracdtt\
intfracdusqrt1+u^2&=intfracdtt\
operatornamearsinh(u)&=log t +c\
u&=sinh(log t+c)
endalign$$
and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.
answered 1 hour ago
mrtaurho
1,190320
add a comment |Â
up vote
1
down vote
By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes
$$fracdudt=frac1tsqrt1+u^2$$
which can be solved by separation of variables. To be exact
$$beginalign
fracdudt&=frac1tsqrt1+u^2\
fracdusqrt1+u^2&=fracdtt\
intfracdusqrt1+u^2&=intfracdtt\
operatornamearsinh(u)&=log t +c\
u&=sinh(log t+c)
endalign$$
and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.
answered 1 hour ago
mrtaurho
1,190320
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes
$$fracdudt=frac1tsqrt1+u^2$$
which can be solved by separation of variables. To be exact
$$beginalign
fracdudt&=frac1tsqrt1+u^2\
fracdusqrt1+u^2&=fracdtt\
intfracdusqrt1+u^2&=intfracdtt\
operatornamearsinh(u)&=log t +c\
u&=sinh(log t+c)
endalign$$
and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.
answered 1 hour ago
mrtaurho
1,190320
By using the substitution $displaystyle u=frac yt$ - and therefore $displaystyle fracdudt=fracty'-yt^2$ - the differential equation becomes
$$fracdudt=frac1tsqrt1+u^2$$
which can be solved by separation of variables. To be exact
$$beginalign
fracdudt&=frac1tsqrt1+u^2\
fracdusqrt1+u^2&=fracdtt\
intfracdusqrt1+u^2&=intfracdtt\
operatornamearsinh(u)&=log t +c\
u&=sinh(log t+c)
endalign$$
and therefore by resubstitution $y(t)=tcdotsinh(log t+c)$.
answered 1 hour ago
mrtaurho
1,190320
edited 1 hour ago
answered 1 hour ago
mrtaurho
1,190320
answered 1 hour ago
mrtaurho
1,190320
answered 1 hour ago
mrtaurho
1,190320
1,190320
add a comment |Â
add a comment |Â
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As a start, switch to $y(-t)$ so that the LHS becomes $-(tcdot y(-t))’$
– b00n heT
1 hour ago