Mixing
Non-homogenous differential equation help
$begingroup$
- a) Given $x = e^u$ and $$x^2frac{d^2y}{dx^2} - 4x frac{dy}{dx} + 6y = 12$$ show that $$frac{d^2y}{du^2} - 5 frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$frac{dx}{du} = e^u = x $$
$$frac{dy}{dx} = frac{dy}{du} frac{du}{dx} = frac{1}{x}frac{dy}{du}$$
$$frac{d^2y}{dx^2} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x}frac{d^2y}{du^2}frac{du}{dx} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}$$
$$x^2left(-frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}right) -4xleft(frac{1}{x}frac{dy}{du}right) + 6y = 12 $$
$$-frac{dy}{du} + frac{d^2y}{du^2} - 4frac{dy}{du} +6y = 12 $$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
part b is where I got stuck:
b) Hence solve the equation $$x^2frac{d^2y}{dx^2} - 4xfrac{dy}{dx} + 6y = 12$$ given that $y(1) = 7$ and $y(2) = 14$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
auxiliary equation: $lambda^2 - 5λ + 6 = 0 $
Hence $lambda = 3, 2$
Complementary function: $$y = Ae^{3u} + Be^{2u}$$
particular integral in the form $y = c$
$$frac{dy}{dx} = 0 = frac{d^2y}{dx^2}$$
$$6C = 12 $$
$$C = 2$$
general solution: $$y = Ae^{3u} + Be^{2u} + 2 $$
where I began to get stuck:
replacing $u = ln x$ means you get left with
$$y = Ax^3 + Bx^2 + 2$$
$$frac{dy}{dx} = 3Ax^2 + 2Bx $$
$$frac{d^2y}{dx^2} = 6Ax + 2B $$
my teacher told me to "substitute this into the solution of the DE and just apply the conditions"
(this is where I think I may me wrong, I think I might have to equate the differential equation to 0 after substituting)
through substituting these values into the differential equation you get:
$$x^2 (6Ax + 2B) - 5(3Ax^2 + 2Bx) + 6 (Ax^3 + Bx^2 + 2) = 12 $$
$$ 6Ax^3 + 2Bx^2 -15Ax^2 - 10Bx + 6Ax^3 + 6Bx^2 + 12 = 12 $$
$$ 12Ax^3 + 8Bx^2 -15Ax^2 - 10Bx = 0 $$
This may resolve very simply, but I didn't want to try without knowing I was doing the correct thing.
ordinary-differential-equations
edited Jan 7 at 9:47
Dylan
12.5k31026
$endgroup$
add a comment |
$begingroup$
- a) Given $x = e^u$ and $$x^2frac{d^2y}{dx^2} - 4x frac{dy}{dx} + 6y = 12$$ show that $$frac{d^2y}{du^2} - 5 frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$frac{dx}{du} = e^u = x $$
$$frac{dy}{dx} = frac{dy}{du} frac{du}{dx} = frac{1}{x}frac{dy}{du}$$
$$frac{d^2y}{dx^2} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x}frac{d^2y}{du^2}frac{du}{dx} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}$$
$$x^2left(-frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}right) -4xleft(frac{1}{x}frac{dy}{du}right) + 6y = 12 $$
$$-frac{dy}{du} + frac{d^2y}{du^2} - 4frac{dy}{du} +6y = 12 $$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
part b is where I got stuck:
b) Hence solve the equation $$x^2frac{d^2y}{dx^2} - 4xfrac{dy}{dx} + 6y = 12$$ given that $y(1) = 7$ and $y(2) = 14$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
auxiliary equation: $lambda^2 - 5λ + 6 = 0 $
Hence $lambda = 3, 2$
Complementary function: $$y = Ae^{3u} + Be^{2u}$$
particular integral in the form $y = c$
$$frac{dy}{dx} = 0 = frac{d^2y}{dx^2}$$
$$6C = 12 $$
$$C = 2$$
general solution: $$y = Ae^{3u} + Be^{2u} + 2 $$
where I began to get stuck:
replacing $u = ln x$ means you get left with
$$y = Ax^3 + Bx^2 + 2$$
$$frac{dy}{dx} = 3Ax^2 + 2Bx $$
$$frac{d^2y}{dx^2} = 6Ax + 2B $$
my teacher told me to "substitute this into the solution of the DE and just apply the conditions"
(this is where I think I may me wrong, I think I might have to equate the differential equation to 0 after substituting)
through substituting these values into the differential equation you get:
$$x^2 (6Ax + 2B) - 5(3Ax^2 + 2Bx) + 6 (Ax^3 + Bx^2 + 2) = 12 $$
$$ 6Ax^3 + 2Bx^2 -15Ax^2 - 10Bx + 6Ax^3 + 6Bx^2 + 12 = 12 $$
$$ 12Ax^3 + 8Bx^2 -15Ax^2 - 10Bx = 0 $$
This may resolve very simply, but I didn't want to try without knowing I was doing the correct thing.
ordinary-differential-equations
edited Jan 7 at 9:47
Dylan
12.5k31026
$endgroup$
$begingroup$
This is way easier than you think. Your general solution $y(x)=Ax^3+Bx^2+2$ is correct. Now, using the boundary conditions, you obtain: $$7=y(1)=Acdot 1^3+Bcdot 1^2+2=A+B+2$$ $$14=y(2)=Acdot 2^3+Bcdot 2^2+2=8A+4B+2$$ Solve for $A$ and $B$, then plug it in to your general solution.
$endgroup$
– projectilemotion
Jan 6 at 21:16
$begingroup$
I appreciate you trying to use the formatting. Next time, use$or$$for the entire line, and never usexfor multiplication
$endgroup$
– Dylan
Jan 7 at 9:49
add a comment |
$begingroup$
- a) Given $x = e^u$ and $$x^2frac{d^2y}{dx^2} - 4x frac{dy}{dx} + 6y = 12$$ show that $$frac{d^2y}{du^2} - 5 frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$frac{dx}{du} = e^u = x $$
$$frac{dy}{dx} = frac{dy}{du} frac{du}{dx} = frac{1}{x}frac{dy}{du}$$
$$frac{d^2y}{dx^2} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x}frac{d^2y}{du^2}frac{du}{dx} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}$$
$$x^2left(-frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}right) -4xleft(frac{1}{x}frac{dy}{du}right) + 6y = 12 $$
$$-frac{dy}{du} + frac{d^2y}{du^2} - 4frac{dy}{du} +6y = 12 $$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
part b is where I got stuck:
b) Hence solve the equation $$x^2frac{d^2y}{dx^2} - 4xfrac{dy}{dx} + 6y = 12$$ given that $y(1) = 7$ and $y(2) = 14$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
auxiliary equation: $lambda^2 - 5λ + 6 = 0 $
Hence $lambda = 3, 2$
Complementary function: $$y = Ae^{3u} + Be^{2u}$$
particular integral in the form $y = c$
$$frac{dy}{dx} = 0 = frac{d^2y}{dx^2}$$
$$6C = 12 $$
$$C = 2$$
general solution: $$y = Ae^{3u} + Be^{2u} + 2 $$
where I began to get stuck:
replacing $u = ln x$ means you get left with
$$y = Ax^3 + Bx^2 + 2$$
$$frac{dy}{dx} = 3Ax^2 + 2Bx $$
$$frac{d^2y}{dx^2} = 6Ax + 2B $$
my teacher told me to "substitute this into the solution of the DE and just apply the conditions"
(this is where I think I may me wrong, I think I might have to equate the differential equation to 0 after substituting)
through substituting these values into the differential equation you get:
$$x^2 (6Ax + 2B) - 5(3Ax^2 + 2Bx) + 6 (Ax^3 + Bx^2 + 2) = 12 $$
$$ 6Ax^3 + 2Bx^2 -15Ax^2 - 10Bx + 6Ax^3 + 6Bx^2 + 12 = 12 $$
$$ 12Ax^3 + 8Bx^2 -15Ax^2 - 10Bx = 0 $$
This may resolve very simply, but I didn't want to try without knowing I was doing the correct thing.
ordinary-differential-equations
edited Jan 7 at 9:47
Dylan
12.5k31026
$endgroup$
- a) Given $x = e^u$ and $$x^2frac{d^2y}{dx^2} - 4x frac{dy}{dx} + 6y = 12$$ show that $$frac{d^2y}{du^2} - 5 frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$frac{dx}{du} = e^u = x $$
$$frac{dy}{dx} = frac{dy}{du} frac{du}{dx} = frac{1}{x}frac{dy}{du}$$
$$frac{d^2y}{dx^2} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x}frac{d^2y}{du^2}frac{du}{dx} = -frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}$$
$$x^2left(-frac{1}{x^2}frac{dy}{du} + frac{1}{x^2}frac{d^2y}{du^2}right) -4xleft(frac{1}{x}frac{dy}{du}right) + 6y = 12 $$
$$-frac{dy}{du} + frac{d^2y}{du^2} - 4frac{dy}{du} +6y = 12 $$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
part b is where I got stuck:
b) Hence solve the equation $$x^2frac{d^2y}{dx^2} - 4xfrac{dy}{dx} + 6y = 12$$ given that $y(1) = 7$ and $y(2) = 14$
$$frac{d^2y}{du^2} - 5frac{dy}{du} + 6y = 12 $$
auxiliary equation: $lambda^2 - 5λ + 6 = 0 $
Hence $lambda = 3, 2$
Complementary function: $$y = Ae^{3u} + Be^{2u}$$
particular integral in the form $y = c$
$$frac{dy}{dx} = 0 = frac{d^2y}{dx^2}$$
$$6C = 12 $$
$$C = 2$$
general solution: $$y = Ae^{3u} + Be^{2u} + 2 $$
where I began to get stuck:
replacing $u = ln x$ means you get left with
$$y = Ax^3 + Bx^2 + 2$$
$$frac{dy}{dx} = 3Ax^2 + 2Bx $$
$$frac{d^2y}{dx^2} = 6Ax + 2B $$
my teacher told me to "substitute this into the solution of the DE and just apply the conditions"
(this is where I think I may me wrong, I think I might have to equate the differential equation to 0 after substituting)
through substituting these values into the differential equation you get:
$$x^2 (6Ax + 2B) - 5(3Ax^2 + 2Bx) + 6 (Ax^3 + Bx^2 + 2) = 12 $$
$$ 6Ax^3 + 2Bx^2 -15Ax^2 - 10Bx + 6Ax^3 + 6Bx^2 + 12 = 12 $$
$$ 12Ax^3 + 8Bx^2 -15Ax^2 - 10Bx = 0 $$
This may resolve very simply, but I didn't want to try without knowing I was doing the correct thing.
ordinary-differential-equations
ordinary-differential-equations
edited Jan 7 at 9:47
Dylan
12.5k31026
edited Jan 7 at 9:47
Dylan
12.5k31026
edited Jan 7 at 9:47
Dylan
12.5k31026
edited Jan 7 at 9:47
Dylan
12.5k31026
edited Jan 7 at 9:47
Dylan
12.5k31026
12.5k31026
asked Jan 6 at 20:50
user624037
$begingroup$
This is way easier than you think. Your general solution $y(x)=Ax^3+Bx^2+2$ is correct. Now, using the boundary conditions, you obtain: $$7=y(1)=Acdot 1^3+Bcdot 1^2+2=A+B+2$$ $$14=y(2)=Acdot 2^3+Bcdot 2^2+2=8A+4B+2$$ Solve for $A$ and $B$, then plug it in to your general solution.
$endgroup$
– projectilemotion
Jan 6 at 21:16
$begingroup$
I appreciate you trying to use the formatting. Next time, use$or$$for the entire line, and never usexfor multiplication
$endgroup$
– Dylan
Jan 7 at 9:49
add a comment |
$begingroup$
This is way easier than you think. Your general solution $y(x)=Ax^3+Bx^2+2$ is correct. Now, using the boundary conditions, you obtain: $$7=y(1)=Acdot 1^3+Bcdot 1^2+2=A+B+2$$ $$14=y(2)=Acdot 2^3+Bcdot 2^2+2=8A+4B+2$$ Solve for $A$ and $B$, then plug it in to your general solution.
$endgroup$
– projectilemotion
Jan 6 at 21:16
$begingroup$
I appreciate you trying to use the formatting. Next time, use$or$$for the entire line, and never usexfor multiplication
$endgroup$
– Dylan
Jan 7 at 9:49
$begingroup$
This is way easier than you think. Your general solution $y(x)=Ax^3+Bx^2+2$ is correct. Now, using the boundary conditions, you obtain: $$7=y(1)=Acdot 1^3+Bcdot 1^2+2=A+B+2$$ $$14=y(2)=Acdot 2^3+Bcdot 2^2+2=8A+4B+2$$ Solve for $A$ and $B$, then plug it in to your general solution.
$endgroup$
– projectilemotion
Jan 6 at 21:16
$begingroup$
This is way easier than you think. Your general solution $y(x)=Ax^3+Bx^2+2$ is correct. Now, using the boundary conditions, you obtain: $$7=y(1)=Acdot 1^3+Bcdot 1^2+2=A+B+2$$ $$14=y(2)=Acdot 2^3+Bcdot 2^2+2=8A+4B+2$$ Solve for $A$ and $B$, then plug it in to your general solution.
$endgroup$
– projectilemotion
Jan 6 at 21:16
$begingroup$
I appreciate you trying to use the formatting. Next time, use
$ or $$ for the entire line, and never use x for multiplication$endgroup$
– Dylan
Jan 7 at 9:49
$begingroup$
I appreciate you trying to use the formatting. Next time, use
$ or $$ for the entire line, and never use x for multiplication$endgroup$
– Dylan
Jan 7 at 9:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
begin{align}
y(1) &= A + B + 2 = 7 \
y(2) &= 8A + 4B + 2 = 14
end{align}
or
begin{align}
A + B &= 5\
8A + 4B &= 12
end{align}
Solving the above system gives $A=-2$, $B=7$
answered Jan 7 at 9:56
DylanDylan
12.5k31026
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
begin{align}
y(1) &= A + B + 2 = 7 \
y(2) &= 8A + 4B + 2 = 14
end{align}
or
begin{align}
A + B &= 5\
8A + 4B &= 12
end{align}
Solving the above system gives $A=-2$, $B=7$
answered Jan 7 at 9:56
DylanDylan
12.5k31026
$endgroup$
add a comment |
$begingroup$
You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
begin{align}
y(1) &= A + B + 2 = 7 \
y(2) &= 8A + 4B + 2 = 14
end{align}
or
begin{align}
A + B &= 5\
8A + 4B &= 12
end{align}
Solving the above system gives $A=-2$, $B=7$
answered Jan 7 at 9:56
DylanDylan
12.5k31026
$endgroup$
add a comment |
$begingroup$
You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
begin{align}
y(1) &= A + B + 2 = 7 \
y(2) &= 8A + 4B + 2 = 14
end{align}
or
begin{align}
A + B &= 5\
8A + 4B &= 12
end{align}
Solving the above system gives $A=-2$, $B=7$
answered Jan 7 at 9:56
DylanDylan
12.5k31026
$endgroup$
You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
begin{align}
y(1) &= A + B + 2 = 7 \
y(2) &= 8A + 4B + 2 = 14
end{align}
or
begin{align}
A + B &= 5\
8A + 4B &= 12
end{align}
Solving the above system gives $A=-2$, $B=7$
answered Jan 7 at 9:56
DylanDylan
12.5k31026
answered Jan 7 at 9:56
DylanDylan
12.5k31026
answered Jan 7 at 9:56
DylanDylan
12.5k31026
answered Jan 7 at 9:56
DylanDylan
12.5k31026
12.5k31026
add a comment |
add a comment |
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$begingroup$
This is way easier than you think. Your general solution $y(x)=Ax^3+Bx^2+2$ is correct. Now, using the boundary conditions, you obtain: $$7=y(1)=Acdot 1^3+Bcdot 1^2+2=A+B+2$$ $$14=y(2)=Acdot 2^3+Bcdot 2^2+2=8A+4B+2$$ Solve for $A$ and $B$, then plug it in to your general solution.
$endgroup$
– projectilemotion
Jan 6 at 21:16
$begingroup$
I appreciate you trying to use the formatting. Next time, use
$or$$for the entire line, and never usexfor multiplication$endgroup$
– Dylan
Jan 7 at 9:49