A calculus problem by Rocco Dalto

Calculus Level pending

{ d x d t = 3 x − 2 y − 3 z d y d t = x + 2 y − 3 z d z d t = x − z \begin{cases} \dfrac{dx}{dt} = 3x - 2y - 3z \\ \dfrac{dy}{dt} = x + 2y - 3z \\ \dfrac{dz}{dt} = x - z \end{cases}

The general solution to the above differential system is as follows:

{ x ( t ) = c 1 x 1 ( t ) + c 2 x 2 ( t ) + c 3 x 3 ( t ) y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + c 3 y 3 ( t ) z ( t ) = c 1 z 1 ( t ) + c 2 z 2 ( t ) + c 3 z 3 ( t ) \begin{cases} x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t) + c_{3}x_{3}(t) \\ y(t) = c_{1}y_{1}(t) + c_{2} y_{2}(t) + c_{3}y_{3}(t) \\ z(t) = c_{1}z_{1}(t) + c_{2}z_{2}(t) + c_{3}z_{3}(t) \end{cases} where ∣ x j ( t ) y j ( t ) z j ( t ) ∣ = ∣ A j B j C j ∣ ∗ e λ j t \begin{vmatrix}{x_{j}(t)} \\{y_{j}(t)} \\ {z_{j}(t)} \\ \end{vmatrix} = \begin{vmatrix}{A_{j}} \\{B_{j}} \\ {C_{j}} \\ \end{vmatrix} * e^{\lambda_{j} t}

for ( 1 < = j < = 3 ) (1 <= j <= 3) .

If C 1 = 1 , C 2 = 1 , C 3 = 1 C_{1} = 1, \: C_{2} = 1, \: C_{3} = 1 and x ( 0 ) = 1 x(0) = 1 , y ( 0 ) = 2 y(0) = 2 , and z ( 0 ) = 3 z(0) = 3 . Find c 1 + c 2 + c 3 c_{1} + c_{2} + c_{3} .

H i n t : Hint: The eigenvalues are λ 1 ∈ R \lambda_{1} \in \mathbb{R} and λ 2 , λ 3 = a ± b i \lambda_{2}, \lambda_{3} = a \pm bi .


The answer is -3.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
May 13, 2017

Let x ( t ) = A ∗ e λ t x(t) = A * e^{\lambda t} y ( t ) = B ∗ e λ t y(t) = B * e^{\lambda t} z ( t ) = C ∗ e λ t z(t) = C * e^{\lambda t}

⟹ d x d t = λ ∗ A ∗ e λ t , d y d t = λ ∗ B ∗ e λ t , d x d t = λ ∗ C ∗ e λ t \implies \dfrac{dx}{dt} = \lambda * A * e^{\lambda t}, \: \dfrac{dy}{dt} = \lambda * B * e^{\lambda t}, \: \dfrac{dx}{dt} = \lambda * C * e^{\lambda t}

Replacing the above in the differential system we obtain the system:

[ 3 − λ − 2 − 3 1 2 − λ − 3 1 0 − 1 − λ ] ∗ ∣ A B C ∣ = ∣ 0 0 0 ∣ \begin{bmatrix}{3 - \lambda} && {-2} && {-3}\\ {1} && {2 - \lambda} && {-3}\\ {1} && {0} && {-1 - \lambda} \end{bmatrix} * \begin{vmatrix}{A} \\{B} \\ {C} \\ \end{vmatrix} = \begin{vmatrix}{0} \\{0} \\ {0} \\ \end{vmatrix}

We want a nontrivial solution so let det ⁡ ( [ 3 − λ − 2 − 3 1 2 − λ − 3 1 0 − 1 − λ ] ) = 0 \det(\begin{bmatrix}{3 - \lambda} && {-2} && {-3}\\ {1} && {2 - \lambda} && {-3}\\ {1} && {0} && {-1 - \lambda} \end{bmatrix}) = 0

⟹ \implies

λ 3 − 4 λ 2 + 6 λ − 4 = ( λ − 2 ) ∗ ( λ 2 − 2 λ + 2 ) = 0 ⟹ λ = 2 , 1 ± i \lambda^3 - 4\lambda^2 + 6\lambda - 4 = (\lambda - 2) * (\lambda^2 - 2\lambda + 2) = 0 \implies \lambda = 2, \: 1 \pm i

For λ = 2 \lambda = 2 we obtain the system:

[ 1 − 2 − 3 0 1 0 − 3 0 1 0 − 3 0 ] \left[ \begin{array}{ccc|c} 1 & -2 & -3 & 0\\ 1 & 0 & -3 & 0\\ 1 & 0 & -3 & 0 \\ \ \end{array} \right]

Using row operations we obtain:

B 1 = 0 B_{1} = 0 , A 1 − 3 C 1 = 0 A_{1} - 3 C_{1} = 0

Letting C 1 = 1 ⟹ A 1 = 3 ⟹ C_{1} = 1 \implies A_{1} = 3 \implies

x 1 ( t ) = 3 e 2 t x_{1}(t) = 3 e^{2t} y 1 ( t ) = 0 y_{1}(t) = 0 z 1 ( t ) = e t z_{1}(t) = e^{t}

For λ = 1 + i \lambda = 1 + i we obtain the system:

[ 2 − i − 2 − 3 0 1 1 − i − 3 0 1 0 − 2 − i 0 ] \left[ \begin{array}{ccc|c} 2 - i & -2 & -3 & 0\\ 1 & 1 - i & -3 & 0\\ 1 & 0 & -2 - i & 0 \\ \ \end{array} \right]

Using row operations we obtain:

A 2 − ( 2 + i ) C 2 = 0 A_{2} - (2 + i) C_{2} = 0 , B 2 − C 2 = 0 B_{2} - C_{2} = 0

Letting C 2 = 1 ⟹ B 2 = 1 ⟹ A 2 = 2 + i C_{2} = 1 \implies B_{2} = 1 \implies A_{2} = 2 + i

∣ A 2 B 2 C 2 ∣ = ∣ 2 + i 1 1 ∣ \begin{vmatrix}{A_{2}} \\{B_{2}} \\ {C_{2}} \\ \end{vmatrix} = \begin{vmatrix}{2 + i} \\{1} \\ {1} \\ \end{vmatrix}

Although it is not needed, for λ = 1 − i \lambda = 1 - i we obtain:

∣ A 3 B 3 C 3 ∣ = ∣ 2 − i 1 1 ∣ \begin{vmatrix}{A_{3}} \\{B_{3}} \\ {C_{3}} \\ \end{vmatrix} = \begin{vmatrix}{2 - i} \\{1} \\ {1} \\ \end{vmatrix}

λ = 1 + i ⟹ \lambda = 1 + i \implies

∣ 2 + i 1 1 ∣ = ∣ 2 1 1 ∣ + ∣ 1 0 0 ∣ i ⟹ \begin{vmatrix}{2 + i} \\{1} \\ {1} \\ \end{vmatrix} = \begin{vmatrix}{2} \\{1} \\ {1} \\ \end{vmatrix} + \begin{vmatrix}{1} \\{0} \\ {0} \\ \end{vmatrix} i \implies

V ⃗ = c 2 e t ∗ ( ∣ 2 1 1 ∣ c o s ( t ) − ∣ 1 0 0 ∣ s i n ( t ) ) + c 3 e t ∗ ( ∣ 2 1 1 ∣ s i n ( t ) + ∣ 1 0 0 ∣ c o s ( t ) ) = \vec{V} = c_{2} e^{t} * ( \begin{vmatrix}{2} \\{1} \\ {1} \\ \end{vmatrix} cos(t) - \begin{vmatrix}{1} \\{0} \\ {0} \\ \end{vmatrix} sin(t)) + c_{3} e^{t} * ( \begin{vmatrix}{2} \\{1} \\ {1} \\ \end{vmatrix} sin(t) + \begin{vmatrix}{1} \\{0} \\ {0} \\ \end{vmatrix} cos(t)) =

c 2 e t ∗ ( ∣ 2 c o s ( t ) − s i n ( t ) c o s ( t ) c o s ( t ) ∣ + c 3 e t ∗ ∣ 2 s i n ( t ) + c o s ( t ) s i n ( t ) s i n ( t ) ∣ c_{2} e^{t} * (\begin{vmatrix}{2 cos(t) - sin(t)} \\{cos(t)} \\ {cos(t)} \\ \end{vmatrix} + c_{3} e^{t} * \begin{vmatrix}{2 sin(t) + cos(t)} \\{sin(t)} \\ {sin(t)} \\ \end{vmatrix}

∴ \therefore The general solution is:

{ x ( t ) = 3 ∗ c 1 ∗ e 2 t + e t [ c 2 ∗ ( 2 c o s ( t ) − s i n ( t ) ) + c 3 ∗ ( 2 s i n ( t ) + c o s ( t ) ) ] y ( t ) = e t ( c 2 ∗ c o s ( t ) + c 3 ∗ s i n ( t ) ) z ( t ) = c 1 ∗ e 2 t + e t ( c 2 ∗ c o s ( t ) + c 3 ∗ s i n ( t ) ) \begin{cases} x(t) = 3 * c_{1} * e^{2t} + e^{t} [c_{2} * (2 cos(t) - sin(t)) + c_{3} * (2 sin(t) + cos(t))] \\ y(t) = e^{t} (c_{2} * cos(t) + c_{3} * sin(t)) \\ z(t) = c_{1} * e^{2t} + e^{t} (c_{2} * cos(t) + c_{3} * sin(t)) \end{cases}

For x ( 0 ) = 1 , y ( 0 ) = 2 , z ( 0 ) = 3 ⟹ x(0) = 1, \: y(0) = 2, \: z(0) = 3 \implies

3 ∗ c 1 + 2 ∗ c 2 + c 3 = 1 3 * c_{1} + 2 * c_{2} + c_{3} = 1 c 2 = 2 c_{2} = 2 c 1 + c 2 = 3 c_{1} + c_{2} = 3

⟹ c 2 = 2 ⟹ c 1 = 1 ⟹ c 3 = − 6 ⟹ \implies c_{2} = 2 \implies c_{1} = 1 \implies c_{3} = -6 \implies

c 1 + c 2 + c 3 = − 3 c_{1} + c_{2} + c_{3} = \boxed{-3}

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