Solving a differential equation with a linear solution and initial conditions
$begingroup$
The problem reads:
The solution of a certain differential equation is of the form
$$y(t)= aexp(5t) + bexp(8t)$$
where $a$ and $b$ are constants.
The solution has initial conditions $y(0)=5$ and $y'(0)=5$
Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................
What I did was solve using the initial conditions and I found that
$a + b = 5 $
and
$
5a + 8b =5. $
Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.
ordinary-differential-equations
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
$endgroup$
add a comment |
$begingroup$
The problem reads:
The solution of a certain differential equation is of the form
$$y(t)= aexp(5t) + bexp(8t)$$
where $a$ and $b$ are constants.
The solution has initial conditions $y(0)=5$ and $y'(0)=5$
Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................
What I did was solve using the initial conditions and I found that
$a + b = 5 $
and
$
5a + 8b =5. $
Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.
ordinary-differential-equations
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
$endgroup$
2
$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09
add a comment |
$begingroup$
The problem reads:
The solution of a certain differential equation is of the form
$$y(t)= aexp(5t) + bexp(8t)$$
where $a$ and $b$ are constants.
The solution has initial conditions $y(0)=5$ and $y'(0)=5$
Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................
What I did was solve using the initial conditions and I found that
$a + b = 5 $
and
$
5a + 8b =5. $
Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.
ordinary-differential-equations
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
$endgroup$
The problem reads:
The solution of a certain differential equation is of the form
$$y(t)= aexp(5t) + bexp(8t)$$
where $a$ and $b$ are constants.
The solution has initial conditions $y(0)=5$ and $y'(0)=5$
Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................
What I did was solve using the initial conditions and I found that
$a + b = 5 $
and
$
5a + 8b =5. $
Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.
ordinary-differential-equations
ordinary-differential-equations
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
edited Sep 4 '16 at 0:36
ForgotALot
3,6271816
3,6271816
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
asked Sep 4 '16 at 0:04
J McDowellJ McDowell
192
192
2
$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09
add a comment |
2
$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09
2
2
$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09
$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're absolutely right.
If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.
Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.
To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.
Personally I would use matrix algebra, but it's up to you.
$$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
left(begin{array}{c} a \ b end{array}right)=
left(begin{array}{c} 5 \ 5 end{array}right) $$
The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives
$$
left(begin{array}{c} a \ b end{array}right)=
frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
left(begin{array}{c} 5 \ 5 end{array}right) $$
Expanding the right hand side gives
begin{eqnarray*}
a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
end{eqnarray*}
Your final solution is then
$$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
$endgroup$
add a comment |
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$begingroup$
You're absolutely right.
If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.
Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.
To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.
Personally I would use matrix algebra, but it's up to you.
$$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
left(begin{array}{c} a \ b end{array}right)=
left(begin{array}{c} 5 \ 5 end{array}right) $$
The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives
$$
left(begin{array}{c} a \ b end{array}right)=
frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
left(begin{array}{c} 5 \ 5 end{array}right) $$
Expanding the right hand side gives
begin{eqnarray*}
a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
end{eqnarray*}
Your final solution is then
$$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
$endgroup$
add a comment |
$begingroup$
You're absolutely right.
If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.
Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.
To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.
Personally I would use matrix algebra, but it's up to you.
$$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
left(begin{array}{c} a \ b end{array}right)=
left(begin{array}{c} 5 \ 5 end{array}right) $$
The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives
$$
left(begin{array}{c} a \ b end{array}right)=
frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
left(begin{array}{c} 5 \ 5 end{array}right) $$
Expanding the right hand side gives
begin{eqnarray*}
a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
end{eqnarray*}
Your final solution is then
$$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
$endgroup$
add a comment |
$begingroup$
You're absolutely right.
If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.
Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.
To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.
Personally I would use matrix algebra, but it's up to you.
$$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
left(begin{array}{c} a \ b end{array}right)=
left(begin{array}{c} 5 \ 5 end{array}right) $$
The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives
$$
left(begin{array}{c} a \ b end{array}right)=
frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
left(begin{array}{c} 5 \ 5 end{array}right) $$
Expanding the right hand side gives
begin{eqnarray*}
a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
end{eqnarray*}
Your final solution is then
$$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
$endgroup$
You're absolutely right.
If $y(t) = amathrm e^{5t} + bmathrm e^{8t}$ then $y'(t) = 5amathrm e^{5t} + 8bmathrm e^{8t}$.
Hence $y(0) = amathrm e^0 + bmathrm e^0 = a+b$ and $y'(0) = 5amathrm e^0 + 8bmathrm e^0 = 5a+8b$.
To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.
Personally I would use matrix algebra, but it's up to you.
$$left(begin{array}{cc} 1 & 1 \ 5 & 8 end{array}right)
left(begin{array}{c} a \ b end{array}right)=
left(begin{array}{c} 5 \ 5 end{array}right) $$
The two-by-two matrix on the right has determinant $1times 8 - 5times 1 = 3 neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives
$$
left(begin{array}{c} a \ b end{array}right)=
frac{1}{3}left(begin{array}{cc} 8 & -1 \ -5 & 1 end{array}right)
left(begin{array}{c} 5 \ 5 end{array}right) $$
Expanding the right hand side gives
begin{eqnarray*}
a &=& frac{1}{3}(8times 5 - 1times 5) &=& frac{35}{3} \ \
b &=& frac{1}{3}(-5times 5 + 1 times 5) &=&-frac{20}{3}
end{eqnarray*}
Your final solution is then
$$boxed{y(t) = tfrac{35}{3}mathrm e^{5t} - tfrac{20}{3}mathrm e^{8t}} $$
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
answered Sep 4 '16 at 1:02
Fly by NightFly by Night
26.1k32978
26.1k32978
add a comment |
add a comment |
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$begingroup$
You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$
$endgroup$
– mfl
Sep 4 '16 at 0:09