Mixing
Solving system of ODE with initial value problem (IVP)
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I have a question about a system of ODE.
If we have:
$fracdxdt=x+2y$
$fracdydt=3x+2y$
with $x(0)=6$ and $y(0)=4$,
how come the solution to the IVP is:
$x(t)=4e^4t+2e^-t$
$y(t)=6e^4t-2e^-t$
I tried doing integral by separating the variable but I didn't get that solution. That example & solution are from my numerical method book, please see the attached image
example & solution
differential-equations differential
asked 1 hour ago
JIM BOY
183
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JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
favorite
I have a question about a system of ODE.
If we have:
$fracdxdt=x+2y$
$fracdydt=3x+2y$
with $x(0)=6$ and $y(0)=4$,
how come the solution to the IVP is:
$x(t)=4e^4t+2e^-t$
$y(t)=6e^4t-2e^-t$
I tried doing integral by separating the variable but I didn't get that solution. That example & solution are from my numerical method book, please see the attached image
example & solution
differential-equations differential
asked 1 hour ago
JIM BOY
183
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What was your solution?
– Neo
1 hour ago
$x(t)+2y(t)=14e^t$ and $3x(t)+2y(t)=26e^t$ I cannot get $e^4t$ or $e^-t$
– JIM BOY
1 hour ago
How did you go about getting these numbers?
– Neo
1 hour ago
For example for $dx/dt$, I separate the variable and use integral, that gives me $ln|x+2y|=t+c_1$. Then I changed the form to $e$, which gives me $x+2y=e^t+c_1$. Then it becomes $x+2y=e^tec_1$, where $e^c_1=C_1$ (a constant).
– JIM BOY
1 hour ago
That is incorrect, as the derivative of the equation does not match what you started with.
– Neo
1 hour ago
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have a question about a system of ODE.
If we have:
$fracdxdt=x+2y$
$fracdydt=3x+2y$
with $x(0)=6$ and $y(0)=4$,
how come the solution to the IVP is:
$x(t)=4e^4t+2e^-t$
$y(t)=6e^4t-2e^-t$
I tried doing integral by separating the variable but I didn't get that solution. That example & solution are from my numerical method book, please see the attached image
example & solution
differential-equations differential
asked 1 hour ago
JIM BOY
183
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a question about a system of ODE.
If we have:
$fracdxdt=x+2y$
$fracdydt=3x+2y$
with $x(0)=6$ and $y(0)=4$,
how come the solution to the IVP is:
$x(t)=4e^4t+2e^-t$
$y(t)=6e^4t-2e^-t$
I tried doing integral by separating the variable but I didn't get that solution. That example & solution are from my numerical method book, please see the attached image
example & solution
differential-equations differential
differential-equations differential
asked 1 hour ago
JIM BOY
183
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JIM BOY
183
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JIM BOY
183
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JIM BOY
183
asked 1 hour ago
JIM BOY
183
183
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
JIM BOY is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What was your solution?
– Neo
1 hour ago
$x(t)+2y(t)=14e^t$ and $3x(t)+2y(t)=26e^t$ I cannot get $e^4t$ or $e^-t$
– JIM BOY
1 hour ago
How did you go about getting these numbers?
– Neo
1 hour ago
For example for $dx/dt$, I separate the variable and use integral, that gives me $ln|x+2y|=t+c_1$. Then I changed the form to $e$, which gives me $x+2y=e^t+c_1$. Then it becomes $x+2y=e^tec_1$, where $e^c_1=C_1$ (a constant).
– JIM BOY
1 hour ago
That is incorrect, as the derivative of the equation does not match what you started with.
– Neo
1 hour ago
 |Â
show 1 more comment
What was your solution?
– Neo
1 hour ago
$x(t)+2y(t)=14e^t$ and $3x(t)+2y(t)=26e^t$ I cannot get $e^4t$ or $e^-t$
– JIM BOY
1 hour ago
How did you go about getting these numbers?
– Neo
1 hour ago
For example for $dx/dt$, I separate the variable and use integral, that gives me $ln|x+2y|=t+c_1$. Then I changed the form to $e$, which gives me $x+2y=e^t+c_1$. Then it becomes $x+2y=e^tec_1$, where $e^c_1=C_1$ (a constant).
– JIM BOY
1 hour ago
That is incorrect, as the derivative of the equation does not match what you started with.
– Neo
1 hour ago
What was your solution?
– Neo
1 hour ago
What was your solution?
– Neo
1 hour ago
$x(t)+2y(t)=14e^t$ and $3x(t)+2y(t)=26e^t$ I cannot get $e^4t$ or $e^-t$
– JIM BOY
1 hour ago
$x(t)+2y(t)=14e^t$ and $3x(t)+2y(t)=26e^t$ I cannot get $e^4t$ or $e^-t$
– JIM BOY
1 hour ago
How did you go about getting these numbers?
– Neo
1 hour ago
How did you go about getting these numbers?
– Neo
1 hour ago
For example for $dx/dt$, I separate the variable and use integral, that gives me $ln|x+2y|=t+c_1$. Then I changed the form to $e$, which gives me $x+2y=e^t+c_1$. Then it becomes $x+2y=e^tec_1$, where $e^c_1=C_1$ (a constant).
– JIM BOY
1 hour ago
For example for $dx/dt$, I separate the variable and use integral, that gives me $ln|x+2y|=t+c_1$. Then I changed the form to $e$, which gives me $x+2y=e^t+c_1$. Then it becomes $x+2y=e^tec_1$, where $e^c_1=C_1$ (a constant).
– JIM BOY
1 hour ago
That is incorrect, as the derivative of the equation does not match what you started with.
– Neo
1 hour ago
That is incorrect, as the derivative of the equation does not match what you started with.
– Neo
1 hour ago
 |Â
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
take laplace transform on both sides of both differential equations to get
$(s-1)X(s)=2Y(s)+6 $ ;
$(s-2)Y(s)=3X(s)+4\$ respectively
solve for $X(s)$ and $Y(s)$ like algebric equations to give ;
$Y(s)=dfrac4s+14(s+1)(s-4)implies y(t)=-2e^-t+6e^4t$
$X(s)=dfrac6s-4(s+1)(s-4)implies x(t)=2e^-t+4e^4t$
answered 58 mins ago
veeresh pandey
886316
1
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
add a comment |Â
up vote
0
down vote
Hint:
One possibility is to eliminate one of the unknowns.
$$x'=x+2yimplies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$
hence
$$x''-3x'-4x=0.$$
answered 1 hour ago
Yves Daoust
117k667213
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
add a comment |Â
up vote
0
down vote
Hint: With $$y=fracx'-x2$$ we get
$$y'=3x+frac32(x'-x)$$ and we get $$y'=fracx''-x'2$$ so we obtain
$$fracx''-x'2=3x+frac32(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^lambda t$$
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
add a comment |Â
up vote
0
down vote
Let us consider your system of ODE:
$$fracdxdt=x+2y\fracdydt=3x+2y$$
It follows that
$$fracdxdt+fracdydt=4x+4yrightarrow fracd(x+y)dt=4(x+y)rightarrow fracd(x+y)x+y=4dt$$
Integration on both sides yields
$$ln(x+y)=4t+Crightarrow x+y=Ce^4trightarrow y=Ce^4t-x$$
Substitution into $fracdxdt$ gives
$$fracdxdt=x+2(Ce^4t-x)rightarrow fracdxdt+x=2Ce^4t$$
Multiply by $e^t$ on both sides:
$$fracdxdte^t+xe^t=2Ce^5trightarrowfracd(xe^t)dt=2Ce^5t$$
Integration on both sides yields
$$xe^t=frac2C5e^5t+Krightarrow x=frac2C5e^4t+Ke^-t$$
Substitution into $y$ gives
$$y=frac3C5e^4t-Ke^-t$$
Apply conditions:
$$x(0)+y(0)=6+4=10rightarrow Ce^4(0)=10rightarrow C=10\x(0)=6rightarrow frac2(10)5e^4(0)+Ke^-(0)=6rightarrow K=2$$
Thus,
$$x(t)=4e^4t+2e^-t\y(t)=6e^4t-2e^-t$$
answered 13 mins ago
Stijn Dietz
3319
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
take laplace transform on both sides of both differential equations to get
$(s-1)X(s)=2Y(s)+6 $ ;
$(s-2)Y(s)=3X(s)+4\$ respectively
solve for $X(s)$ and $Y(s)$ like algebric equations to give ;
$Y(s)=dfrac4s+14(s+1)(s-4)implies y(t)=-2e^-t+6e^4t$
$X(s)=dfrac6s-4(s+1)(s-4)implies x(t)=2e^-t+4e^4t$
answered 58 mins ago
veeresh pandey
886316
1
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
add a comment |Â
up vote
2
down vote
accepted
take laplace transform on both sides of both differential equations to get
$(s-1)X(s)=2Y(s)+6 $ ;
$(s-2)Y(s)=3X(s)+4\$ respectively
solve for $X(s)$ and $Y(s)$ like algebric equations to give ;
$Y(s)=dfrac4s+14(s+1)(s-4)implies y(t)=-2e^-t+6e^4t$
$X(s)=dfrac6s-4(s+1)(s-4)implies x(t)=2e^-t+4e^4t$
answered 58 mins ago
veeresh pandey
886316
1
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
take laplace transform on both sides of both differential equations to get
$(s-1)X(s)=2Y(s)+6 $ ;
$(s-2)Y(s)=3X(s)+4\$ respectively
solve for $X(s)$ and $Y(s)$ like algebric equations to give ;
$Y(s)=dfrac4s+14(s+1)(s-4)implies y(t)=-2e^-t+6e^4t$
$X(s)=dfrac6s-4(s+1)(s-4)implies x(t)=2e^-t+4e^4t$
answered 58 mins ago
veeresh pandey
886316
take laplace transform on both sides of both differential equations to get
$(s-1)X(s)=2Y(s)+6 $ ;
$(s-2)Y(s)=3X(s)+4\$ respectively
solve for $X(s)$ and $Y(s)$ like algebric equations to give ;
$Y(s)=dfrac4s+14(s+1)(s-4)implies y(t)=-2e^-t+6e^4t$
$X(s)=dfrac6s-4(s+1)(s-4)implies x(t)=2e^-t+4e^4t$
answered 58 mins ago
veeresh pandey
886316
edited 48 mins ago
answered 58 mins ago
veeresh pandey
886316
answered 58 mins ago
veeresh pandey
886316
answered 58 mins ago
veeresh pandey
886316
886316
1
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
add a comment |Â
1
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
1
1
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
Thank you! looks like i missed some of the steps to solve ODE with IVP
– JIM BOY
54 mins ago
add a comment |Â
up vote
0
down vote
Hint:
One possibility is to eliminate one of the unknowns.
$$x'=x+2yimplies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$
hence
$$x''-3x'-4x=0.$$
answered 1 hour ago
Yves Daoust
117k667213
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
add a comment |Â
up vote
0
down vote
Hint:
One possibility is to eliminate one of the unknowns.
$$x'=x+2yimplies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$
hence
$$x''-3x'-4x=0.$$
answered 1 hour ago
Yves Daoust
117k667213
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
One possibility is to eliminate one of the unknowns.
$$x'=x+2yimplies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$
hence
$$x''-3x'-4x=0.$$
answered 1 hour ago
Yves Daoust
117k667213
Hint:
One possibility is to eliminate one of the unknowns.
$$x'=x+2yimplies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$
hence
$$x''-3x'-4x=0.$$
answered 1 hour ago
Yves Daoust
117k667213
answered 1 hour ago
Yves Daoust
117k667213
answered 1 hour ago
Yves Daoust
117k667213
answered 1 hour ago
Yves Daoust
117k667213
117k667213
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
add a comment |Â
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
I'm not too sure how to get the e from there
– JIM BOY
1 hour ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
@JIMBOY: review your theory of univariate linear ODE with constant coefficients.
– Yves Daoust
59 mins ago
add a comment |Â
up vote
0
down vote
Hint: With $$y=fracx'-x2$$ we get
$$y'=3x+frac32(x'-x)$$ and we get $$y'=fracx''-x'2$$ so we obtain
$$fracx''-x'2=3x+frac32(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^lambda t$$
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
add a comment |Â
up vote
0
down vote
Hint: With $$y=fracx'-x2$$ we get
$$y'=3x+frac32(x'-x)$$ and we get $$y'=fracx''-x'2$$ so we obtain
$$fracx''-x'2=3x+frac32(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^lambda t$$
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: With $$y=fracx'-x2$$ we get
$$y'=3x+frac32(x'-x)$$ and we get $$y'=fracx''-x'2$$ so we obtain
$$fracx''-x'2=3x+frac32(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^lambda t$$
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
Hint: With $$y=fracx'-x2$$ we get
$$y'=3x+frac32(x'-x)$$ and we get $$y'=fracx''-x'2$$ so we obtain
$$fracx''-x'2=3x+frac32(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^lambda t$$
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
edited 59 mins ago
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
answered 1 hour ago
Dr. Sonnhard Graubner
69.7k32862
69.7k32862
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
add a comment |Â
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
I'm not too sure how to get the $e$ from there
– JIM BOY
1 hour ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
Thanks, I'll go check on the $x=e^lambda t$
– JIM BOY
56 mins ago
add a comment |Â
up vote
0
down vote
Let us consider your system of ODE:
$$fracdxdt=x+2y\fracdydt=3x+2y$$
It follows that
$$fracdxdt+fracdydt=4x+4yrightarrow fracd(x+y)dt=4(x+y)rightarrow fracd(x+y)x+y=4dt$$
Integration on both sides yields
$$ln(x+y)=4t+Crightarrow x+y=Ce^4trightarrow y=Ce^4t-x$$
Substitution into $fracdxdt$ gives
$$fracdxdt=x+2(Ce^4t-x)rightarrow fracdxdt+x=2Ce^4t$$
Multiply by $e^t$ on both sides:
$$fracdxdte^t+xe^t=2Ce^5trightarrowfracd(xe^t)dt=2Ce^5t$$
Integration on both sides yields
$$xe^t=frac2C5e^5t+Krightarrow x=frac2C5e^4t+Ke^-t$$
Substitution into $y$ gives
$$y=frac3C5e^4t-Ke^-t$$
Apply conditions:
$$x(0)+y(0)=6+4=10rightarrow Ce^4(0)=10rightarrow C=10\x(0)=6rightarrow frac2(10)5e^4(0)+Ke^-(0)=6rightarrow K=2$$
Thus,
$$x(t)=4e^4t+2e^-t\y(t)=6e^4t-2e^-t$$
answered 13 mins ago
Stijn Dietz
3319
add a comment |Â
up vote
0
down vote
Let us consider your system of ODE:
$$fracdxdt=x+2y\fracdydt=3x+2y$$
It follows that
$$fracdxdt+fracdydt=4x+4yrightarrow fracd(x+y)dt=4(x+y)rightarrow fracd(x+y)x+y=4dt$$
Integration on both sides yields
$$ln(x+y)=4t+Crightarrow x+y=Ce^4trightarrow y=Ce^4t-x$$
Substitution into $fracdxdt$ gives
$$fracdxdt=x+2(Ce^4t-x)rightarrow fracdxdt+x=2Ce^4t$$
Multiply by $e^t$ on both sides:
$$fracdxdte^t+xe^t=2Ce^5trightarrowfracd(xe^t)dt=2Ce^5t$$
Integration on both sides yields
$$xe^t=frac2C5e^5t+Krightarrow x=frac2C5e^4t+Ke^-t$$
Substitution into $y$ gives
$$y=frac3C5e^4t-Ke^-t$$
Apply conditions:
$$x(0)+y(0)=6+4=10rightarrow Ce^4(0)=10rightarrow C=10\x(0)=6rightarrow frac2(10)5e^4(0)+Ke^-(0)=6rightarrow K=2$$
Thus,
$$x(t)=4e^4t+2e^-t\y(t)=6e^4t-2e^-t$$
answered 13 mins ago
Stijn Dietz
3319
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let us consider your system of ODE:
$$fracdxdt=x+2y\fracdydt=3x+2y$$
It follows that
$$fracdxdt+fracdydt=4x+4yrightarrow fracd(x+y)dt=4(x+y)rightarrow fracd(x+y)x+y=4dt$$
Integration on both sides yields
$$ln(x+y)=4t+Crightarrow x+y=Ce^4trightarrow y=Ce^4t-x$$
Substitution into $fracdxdt$ gives
$$fracdxdt=x+2(Ce^4t-x)rightarrow fracdxdt+x=2Ce^4t$$
Multiply by $e^t$ on both sides:
$$fracdxdte^t+xe^t=2Ce^5trightarrowfracd(xe^t)dt=2Ce^5t$$
Integration on both sides yields
$$xe^t=frac2C5e^5t+Krightarrow x=frac2C5e^4t+Ke^-t$$
Substitution into $y$ gives
$$y=frac3C5e^4t-Ke^-t$$
Apply conditions:
$$x(0)+y(0)=6+4=10rightarrow Ce^4(0)=10rightarrow C=10\x(0)=6rightarrow frac2(10)5e^4(0)+Ke^-(0)=6rightarrow K=2$$
Thus,
$$x(t)=4e^4t+2e^-t\y(t)=6e^4t-2e^-t$$
answered 13 mins ago
Stijn Dietz
3319
Let us consider your system of ODE:
$$fracdxdt=x+2y\fracdydt=3x+2y$$
It follows that
$$fracdxdt+fracdydt=4x+4yrightarrow fracd(x+y)dt=4(x+y)rightarrow fracd(x+y)x+y=4dt$$
Integration on both sides yields
$$ln(x+y)=4t+Crightarrow x+y=Ce^4trightarrow y=Ce^4t-x$$
Substitution into $fracdxdt$ gives
$$fracdxdt=x+2(Ce^4t-x)rightarrow fracdxdt+x=2Ce^4t$$
Multiply by $e^t$ on both sides:
$$fracdxdte^t+xe^t=2Ce^5trightarrowfracd(xe^t)dt=2Ce^5t$$
Integration on both sides yields
$$xe^t=frac2C5e^5t+Krightarrow x=frac2C5e^4t+Ke^-t$$
Substitution into $y$ gives
$$y=frac3C5e^4t-Ke^-t$$
Apply conditions:
$$x(0)+y(0)=6+4=10rightarrow Ce^4(0)=10rightarrow C=10\x(0)=6rightarrow frac2(10)5e^4(0)+Ke^-(0)=6rightarrow K=2$$
Thus,
$$x(t)=4e^4t+2e^-t\y(t)=6e^4t-2e^-t$$
answered 13 mins ago
Stijn Dietz
3319
answered 13 mins ago
Stijn Dietz
3319
answered 13 mins ago
Stijn Dietz
3319
answered 13 mins ago
Stijn Dietz
3319
3319
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add a comment |Â
JIM BOY is a new contributor. Be nice, and check out our Code of Conduct.
JIM BOY is a new contributor. Be nice, and check out our Code of Conduct.
JIM BOY is a new contributor. Be nice, and check out our Code of Conduct.
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What was your solution?
– Neo
1 hour ago
$x(t)+2y(t)=14e^t$ and $3x(t)+2y(t)=26e^t$ I cannot get $e^4t$ or $e^-t$
– JIM BOY
1 hour ago
How did you go about getting these numbers?
– Neo
1 hour ago
For example for $dx/dt$, I separate the variable and use integral, that gives me $ln|x+2y|=t+c_1$. Then I changed the form to $e$, which gives me $x+2y=e^t+c_1$. Then it becomes $x+2y=e^tec_1$, where $e^c_1=C_1$ (a constant).
– JIM BOY
1 hour ago
That is incorrect, as the derivative of the equation does not match what you started with.
– Neo
1 hour ago