ANSWER on Question #80494 – Math – Differential Equations
QUESTION
Using the method of undermined coefficient find the general solution of D.E
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 y e − x + 2 e − x ⋅ x + e − x ⋅ sin x y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 3 y e ^ {- x} + 2 e ^ {- x} \cdot x + e ^ {- x} \cdot \sin x y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 y e − x + 2 e − x ⋅ x + e − x ⋅ sin x SOLUTION
Hint: In the solution, I will proceed from the assumption that the question contained a mistake and the equation actually looks like this:
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 e − x + 2 e − x ⋅ x + e − x ⋅ sin x y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 3 e ^ {- x} + 2 e ^ {- x} \cdot x + e ^ {- x} \cdot \sin x y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 e − x + 2 e − x ⋅ x + e − x ⋅ sin x
As we know from theory, the solution of a differential equation of the form
F ( x , y , y ′ , y ′ ′ , … , y ( n ) ) = f 1 ( x ) + f 2 ( x ) + f 3 ( x ) + … f n ( x ) F \big (x, y, y ^ {\prime}, y ^ {\prime \prime}, \dots , y ^ {(n)} \big) = f _ {1} (x) + f _ {2} (x) + f _ {3} (x) + \dots f _ {n} (x) F ( x , y , y ′ , y ′′ , … , y ( n ) ) = f 1 ( x ) + f 2 ( x ) + f 3 ( x ) + … f n ( x )
consists of such parts:
1) solution of the homogeneous equation
F ( x , y , y ′ , y ′ ′ , … , y ( n ) ) = 0 → y = y h ( x ) F \big (x, y, y ^ {\prime}, y ^ {\prime \prime}, \dots , y ^ {(n)} \big) = 0 \rightarrow y = y _ {h} (x) F ( x , y , y ′ , y ′′ , … , y ( n ) ) = 0 → y = y h ( x )
2) partial solutions of inhomogeneous equations
F ( x , y , y ′ , y ′ ′ , … , y ( n ) ) = f 1 ( x ) → y = y 1 ( x ) F \big (x, y, y ^ {\prime}, y ^ {\prime \prime}, \dots , y ^ {(n)} \big) = f _ {1} (x) \rightarrow y = y _ {1} (x) F ( x , y , y ′ , y ′′ , … , y ( n ) ) = f 1 ( x ) → y = y 1 ( x ) F ( x , y , y ′ , y ′ ′ , … , y ( n ) ) = f 2 ( x ) → y = y 2 ( x ) F \big (x, y, y ^ {\prime}, y ^ {\prime \prime}, \dots , y ^ {(n)} \big) = f _ {2} (x) \rightarrow y = y _ {2} (x) F ( x , y , y ′ , y ′′ , … , y ( n ) ) = f 2 ( x ) → y = y 2 ( x ) F ( x , y , y ′ , y ′ ′ , … , y ( n ) ) = f 3 ( x ) → y = y 3 ( x ) F \big (x, y, y ^ {\prime}, y ^ {\prime \prime}, \dots , y ^ {(n)} \big) = f _ {3} (x) \rightarrow y = y _ {3} (x) F ( x , y , y ′ , y ′′ , … , y ( n ) ) = f 3 ( x ) → y = y 3 ( x ) … … … … … … … … … … … … … … \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots …………………………………… F ( x , y , y ′ , y ′ ′ , … , y ( n ) ) = f n ( x ) → y = y n ( x ) F \big (x, y, y ^ {\prime}, y ^ {\prime \prime}, \dots , y ^ {(n)} \big) = f _ {n} (x) \rightarrow y = y _ {n} (x) F ( x , y , y ′ , y ′′ , … , y ( n ) ) = f n ( x ) → y = y n ( x )
Then,
y s o l = y h ( x ) + y 1 ( x ) + y 2 ( x ) + y 3 ( x ) + ⋯ + y n ( x ) y _ {s o l} = y _ {h} (x) + y _ {1} (x) + y _ {2} (x) + y _ {3} (x) + \dots + y _ {n} (x) y so l = y h ( x ) + y 1 ( x ) + y 2 ( x ) + y 3 ( x ) + ⋯ + y n ( x )
In our case,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 e − x + 2 e − x ⋅ x + e − x ⋅ sin x y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 3 e ^ {- x} + 2 e ^ {- x} \cdot x + e ^ {- x} \cdot \sin x y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 e − x + 2 e − x ⋅ x + e − x ⋅ sin x
1 STEP: We find the solution of the homogeneous equation
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 0 y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 0 y ( I V ) − 2 y ′′′ + 2 y ′′ = 0
The solution will be sought in the form
y = e k x → v → y = e ^ {k x} \rightarrow v \rightarrow y = e k x → v → y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 0 → k 4 ⋅ e k x − 2 k 3 ⋅ e k x + 2 k 2 ⋅ e k x = 0 → e k x ⋅ ( k 4 − 2 k 3 + 2 k 2 ) = 0 → y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 0 \rightarrow k ^ {4} \cdot e ^ {k x} - 2 k ^ {3} \cdot e ^ {k x} + 2 k ^ {2} \cdot e ^ {k x} = 0 \rightarrow e ^ {k x} \cdot (k ^ {4} - 2 k ^ {3} + 2 k ^ {2}) = 0 \rightarrow y ( I V ) − 2 y ′′′ + 2 y ′′ = 0 → k 4 ⋅ e k x − 2 k 3 ⋅ e k x + 2 k 2 ⋅ e k x = 0 → e k x ⋅ ( k 4 − 2 k 3 + 2 k 2 ) = 0 → k 4 − 2 k 3 + 2 k 2 = 0 → k 2 ⋅ ( k 2 − 2 k + 2 ) = 0 → [ k 2 = 0 k 2 − 2 k + 2 = 0 k ^ {4} - 2 k ^ {3} + 2 k ^ {2} = 0 \rightarrow k ^ {2} \cdot (k ^ {2} - 2 k + 2) = 0 \rightarrow \left[\begin{array}{c}k ^ {2} = 0\\k ^ {2} - 2 k + 2 = 0\end{array}\right. k 4 − 2 k 3 + 2 k 2 = 0 → k 2 ⋅ ( k 2 − 2 k + 2 ) = 0 → [ k 2 = 0 k 2 − 2 k + 2 = 0 k 2 = 0 → k 1 , 2 = 0 − m u l t i p l e r o o t k ^ {2} = 0 \rightarrow \boxed {k _ {1, 2} = 0 - m u l t i p l e r o o t} k 2 = 0 → k 1 , 2 = 0 − m u lt i pl eroo t k 2 − 2 k + 2 = 0 → { a = 1 b = − 2 c = 2 → D = b 2 − 4 a c = ( − 2 ) 2 − 4 ⋅ 1 ⋅ 2 = 4 − 8 = − 4 → k ^ {2} - 2 k + 2 = 0 \rightarrow \left\{\begin{array}{l}a = 1\\b = - 2\\c = 2\end{array}\right. \rightarrow D = b ^ {2} - 4 a c = (- 2) ^ {2} - 4 \cdot 1 \cdot 2 = 4 - 8 = - 4 \rightarrow k 2 − 2 k + 2 = 0 → ⎩ ⎨ ⎧ a = 1 b = − 2 c = 2 → D = b 2 − 4 a c = ( − 2 ) 2 − 4 ⋅ 1 ⋅ 2 = 4 − 8 = − 4 → D = − 4 = 2 i → [ k 3 = − b − D 2 a = − ( − 2 ) − 2 i 2 ⋅ 1 = 2 − 2 i 2 = 1 − i k 4 = − b + D 2 a = − ( − 2 ) + 2 i 2 ⋅ 1 = 2 + 2 i 2 = 1 + i \sqrt {D} = \sqrt {- 4} = 2 i \rightarrow \left[\begin{array}{l}k _ {3} = \frac {- b - \sqrt {D}}{2 a} = \frac {- (- 2) - 2 i}{2 \cdot 1} = \frac {2 - 2 i}{2} = 1 - i\\k _ {4} = \frac {- b + \sqrt {D}}{2 a} = \frac {- (- 2) + 2 i}{2 \cdot 1} = \frac {2 + 2 i}{2} = 1 + i\end{array}\right. D = − 4 = 2 i → [ k 3 = 2 a − b − D = 2 ⋅ 1 − ( − 2 ) − 2 i = 2 2 − 2 i = 1 − i k 4 = 2 a − b + D = 2 ⋅ 1 − ( − 2 ) + 2 i = 2 2 + 2 i = 1 + i
Then,
y h ( x ) = e k 1 x ( A + B x ) ⏟ since the root of multiplicity 2 + C 1 e k 3 x + C 2 e k 4 x → y_{h}(x) = \underbrace{e^{k_{1}x}(A + Bx)}_{\text{since the root of multiplicity 2}} + C_{1}e^{k_{3}x} + C_{2}e^{k_{4}x}\rightarrow y h ( x ) = since the root of multiplicity 2 e k 1 x ( A + B x ) + C 1 e k 3 x + C 2 e k 4 x → y h ( x ) = e 0 ⋅ x ( A + B x ) + C 1 e ( 1 + i ) x + C 3 e ( 1 − i ) x ≡ A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x y _ {h} (x) = e ^ {0 \cdot x} (A + B x) + C _ {1} e ^ {(1 + i) x} + C _ {3} e ^ {(1 - i) x} \equiv A + B x + A _ {1} e ^ {x} \cdot \cos x + A _ {2} e ^ {x} \cdot \sin x y h ( x ) = e 0 ⋅ x ( A + B x ) + C 1 e ( 1 + i ) x + C 3 e ( 1 − i ) x ≡ A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x Hint: { e i x = cos x + i ⋅ sin x e − i x = cos x + i ⋅ sin x → C 1 e i x + C 2 e − i x ≡ A 1 cos x + A 2 sin x \text{Hint:} \left\{ \begin{array}{l} e ^ {i x} = \cos x + i \cdot \sin x \\ e ^ {- i x} = \cos x + i \cdot \sin x \end{array} \right. \to C _ {1} e ^ {i x} + C _ {2} e ^ {- i x} \equiv A _ {1} \cos x + A _ {2} \sin x Hint: { e i x = cos x + i ⋅ sin x e − i x = cos x + i ⋅ sin x → C 1 e i x + C 2 e − i x ≡ A 1 cos x + A 2 sin x
Conclusion,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 0 → y h ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x \boxed {y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 0 \rightarrow y _ {h} (x) = A + B x + A _ {1} e ^ {x} \cdot \cos x + A _ {2} e ^ {x} \cdot \sin x} y ( I V ) − 2 y ′′′ + 2 y ′′ = 0 → y h ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x
2 STEP: We find a particular solution of the inhomogeneous equations
a) f 1 ( x ) = 3 e − x f_{1}(x) = 3e^{-x} f 1 ( x ) = 3 e − x
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 e − x y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 3 e ^ {- x} y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 e − x
The solution will be sought in the form
y ( x ) = C e − x → { y ( I V ) = C ⋅ ( − 1 ) 4 ⋅ e − x = C e − x y ′ ′ ′ = C ⋅ ( − 1 ) 3 ⋅ e − x = − C e − x y ′ ′ = C ⋅ ( − 1 ) 2 ⋅ e − x = C e − x y(x) = C e^{-x} \rightarrow \left\{ \begin{array}{l} y^{(IV)} = C \cdot (-1)^4 \cdot e^{-x} = C e^{-x} \\ y''' = C \cdot (-1)^3 \cdot e^{-x} = -C e^{-x} \\ y'' = C \cdot (-1)^2 \cdot e^{-x} = C e^{-x} \end{array} \right. y ( x ) = C e − x → ⎩ ⎨ ⎧ y ( I V ) = C ⋅ ( − 1 ) 4 ⋅ e − x = C e − x y ′′′ = C ⋅ ( − 1 ) 3 ⋅ e − x = − C e − x y ′′ = C ⋅ ( − 1 ) 2 ⋅ e − x = C e − x
Then,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 e − x → C e − x − 2 ⋅ ( − C e − x ) + 2 ⋅ C e − x = 3 e − x → 5 C e − x = 3 e − x ∣ ÷ ( 5 e − x ) → y^{(IV)} - 2 y''' + 2 y'' = 3 e^{-x} \rightarrow C e^{-x} - 2 \cdot (-C e^{-x}) + 2 \cdot C e^{-x} = 3 e^{-x} \rightarrow 5 C e^{-x} = 3 e^{-x} | \div (5 e^{-x}) \rightarrow y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 e − x → C e − x − 2 ⋅ ( − C e − x ) + 2 ⋅ C e − x = 3 e − x → 5 C e − x = 3 e − x ∣ ÷ ( 5 e − x ) → C = 3 e − x 5 e − x = 3 5 C = \frac{3 e^{-x}}{5 e^{-x}} = \frac{3}{5} C = 5 e − x 3 e − x = 5 3
Conclusion,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 e − x → y 1 ( x ) = 3 5 e − x y^{(IV)} - 2 y''' + 2 y'' = 3 e^{-x} \rightarrow y_1(x) = \frac{3}{5} e^{-x} y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 e − x → y 1 ( x ) = 5 3 e − x
b) f 2 ( x ) = 2 e − x ⋅ x f_2(x) = 2 e^{-x} \cdot x f 2 ( x ) = 2 e − x ⋅ x
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 2 e − x ⋅ x y^{(IV)} - 2 y''' + 2 y'' = 2 e^{-x} \cdot x y ( I V ) − 2 y ′′′ + 2 y ′′ = 2 e − x ⋅ x
The solution will be sought in the form
y ( x ) = e − x ⋅ ( A x + B ) → { y ′ = − e − x ⋅ ( A x + B ) + A e − x ≡ e − x ⋅ ( A − B − A x ) y ′ ′ = − e − x ⋅ ( A − B − A x ) − A e − x ≡ e − x ⋅ ( B − 2 A + A x ) y ′ ′ ′ = − e − x ⋅ ( B − 2 A + A x ) + A e − x ≡ e − x ⋅ ( 3 A − B − A x ) y ( I V ) = − e − x ⋅ ( 3 A − B − A x ) − A e − x ≡ e − x ⋅ ( B − 4 A + A x ) y(x) = e^{-x} \cdot (A x + B) \rightarrow \left\{ \begin{array}{l} y' = -e^{-x} \cdot (A x + B) + A e^{-x} \equiv e^{-x} \cdot (A - B - A x) \\ y'' = -e^{-x} \cdot (A - B - A x) - A e^{-x} \equiv e^{-x} \cdot (B - 2 A + A x) \\ y''' = -e^{-x} \cdot (B - 2 A + A x) + A e^{-x} \equiv e^{-x} \cdot (3 A - B - A x) \\ y^{(IV)} = -e^{-x} \cdot (3 A - B - A x) - A e^{-x} \equiv e^{-x} \cdot (B - 4 A + A x) \end{array} \right. y ( x ) = e − x ⋅ ( A x + B ) → ⎩ ⎨ ⎧ y ′ = − e − x ⋅ ( A x + B ) + A e − x ≡ e − x ⋅ ( A − B − A x ) y ′′ = − e − x ⋅ ( A − B − A x ) − A e − x ≡ e − x ⋅ ( B − 2 A + A x ) y ′′′ = − e − x ⋅ ( B − 2 A + A x ) + A e − x ≡ e − x ⋅ ( 3 A − B − A x ) y ( I V ) = − e − x ⋅ ( 3 A − B − A x ) − A e − x ≡ e − x ⋅ ( B − 4 A + A x )
Then,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 2 e − x ⋅ x → y^{(IV)} - 2 y''' + 2 y'' = 2 e^{-x} \cdot x \rightarrow y ( I V ) − 2 y ′′′ + 2 y ′′ = 2 e − x ⋅ x → e − x ⋅ ( B − 4 A + A x ) − 2 ⋅ e − x ⋅ ( 3 A − B − A x ) + 2 ⋅ e − x ⋅ ( B − 2 A + A x ) = e − x ⋅ ( 2 x ) → e^{-x} \cdot (B - 4 A + A x) - 2 \cdot e^{-x} \cdot (3 A - B - A x) + 2 \cdot e^{-x} \cdot (B - 2 A + A x) = e^{-x} \cdot (2 x) \rightarrow e − x ⋅ ( B − 4 A + A x ) − 2 ⋅ e − x ⋅ ( 3 A − B − A x ) + 2 ⋅ e − x ⋅ ( B − 2 A + A x ) = e − x ⋅ ( 2 x ) → e − x ⋅ ( B − 4 A + A x − 6 A + 2 B + 2 A x + 2 B − 4 A + 2 A x ) = e − x ⋅ ( 2 x ) → e^{-x} \cdot (B - 4 A + A x - 6 A + 2 B + 2 A x + 2 B - 4 A + 2 A x) = e^{-x} \cdot (2 x) \rightarrow e − x ⋅ ( B − 4 A + A x − 6 A + 2 B + 2 A x + 2 B − 4 A + 2 A x ) = e − x ⋅ ( 2 x ) → e − x ⋅ ( 5 B − 14 A + 5 A x ) = e − x ⋅ ( 2 x ) → 5 B − 14 A + 5 A x = 2 x → { 5 A = 2 ∣ ÷ ( 5 ) 5 B − 14 A = 0 → e^{-x} \cdot (5 B - 14 A + 5 A x) = e^{-x} \cdot (2 x) \rightarrow 5 B - 14 A + 5 A x = 2 x \rightarrow \left\{ \begin{array}{l} 5 A = 2 | \div (5) \\ 5 B - 14 A = 0 \end{array} \right. \rightarrow e − x ⋅ ( 5 B − 14 A + 5 A x ) = e − x ⋅ ( 2 x ) → 5 B − 14 A + 5 A x = 2 x → { 5 A = 2∣ ÷ ( 5 ) 5 B − 14 A = 0 → { A = 2 5 5 B = 14 ⋅ 2 5 → { A = 2 5 B = 28 25 \left\{ \begin{array}{c} A = \frac{2}{5} \\ 5 B = 14 \cdot \frac{2}{5} \end{array} \right. \rightarrow \boxed{\left\{ \begin{array}{l} A = \frac{2}{5} \\ B = \frac{28}{25} \end{array} \right.} { A = 5 2 5 B = 14 ⋅ 5 2 → { A = 5 2 B = 25 28
Conclusion,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 2 e − x ⋅ x → y 2 ( x ) = e − x ⋅ ( 2 x 5 + 28 25 ) y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 2 e ^ {- x} \cdot x \rightarrow y _ {2} (x) = e ^ {- x} \cdot \left(\frac {2 x}{5} + \frac {2 8}{2 5}\right) y ( I V ) − 2 y ′′′ + 2 y ′′ = 2 e − x ⋅ x → y 2 ( x ) = e − x ⋅ ( 5 2 x + 25 28 )
c) f 3 ( x ) = e − x ⋅ sin x f_{3}(x) = e^{-x}\cdot \sin x f 3 ( x ) = e − x ⋅ sin x
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = e − x ⋅ sin x y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = e ^ {- x} \cdot \sin x y ( I V ) − 2 y ′′′ + 2 y ′′ = e − x ⋅ sin x
The solution will be sought in the form
y ( x ) = A e − x ⋅ sin x + B e − x ⋅ cos x → y (x) = A e ^ {- x} \cdot \sin x + B e ^ {- x} \cdot \cos x \rightarrow y ( x ) = A e − x ⋅ sin x + B e − x ⋅ cos x → y ′ = − A e − x ⋅ sin x + A e − x ⋅ cos x − B e − x ⋅ cos x − B e − x ⋅ sin x → y ^ {\prime} = - A e ^ {- x} \cdot \sin x + A e ^ {- x} \cdot \cos x - B e ^ {- x} \cdot \cos x - B e ^ {- x} \cdot \sin x \rightarrow y ′ = − A e − x ⋅ sin x + A e − x ⋅ cos x − B e − x ⋅ cos x − B e − x ⋅ sin x → y ′ = ( − A − B ) e − x ⋅ sin x + ( A − B ) e − x ⋅ cos x y ^ {\prime} = (- A - B) e ^ {- x} \cdot \sin x + (A - B) e ^ {- x} \cdot \cos x y ′ = ( − A − B ) e − x ⋅ sin x + ( A − B ) e − x ⋅ cos x y ′ ′ = − ( − A − B ) e − x ⋅ sin x + ( − A − B ) e − x ⋅ cos x − ( A − B ) e − x ⋅ cos x − ( A − B ) e − x ⋅ sin x → y ^ {\prime \prime} = - (- A - B) e ^ {- x} \cdot \sin x + (- A - B) e ^ {- x} \cdot \cos x - (A - B) e ^ {- x} \cdot \cos x - (A - B) e ^ {- x} \cdot \sin x \rightarrow y ′′ = − ( − A − B ) e − x ⋅ sin x + ( − A − B ) e − x ⋅ cos x − ( A − B ) e − x ⋅ cos x − ( A − B ) e − x ⋅ sin x → y ′ ′ = ( A + B − A + B ) e − x ⋅ sin x + ( − A − B − A + B ) e − x ⋅ cos x → y ^ {\prime \prime} = (A + B - A + B) e ^ {- x} \cdot \sin x + (- A - B - A + B) e ^ {- x} \cdot \cos x \rightarrow y ′′ = ( A + B − A + B ) e − x ⋅ sin x + ( − A − B − A + B ) e − x ⋅ cos x → y ′ ′ = ( 2 B ) e − x ⋅ sin x + ( − 2 A ) e − x ⋅ cos x y ^ {\prime \prime} = (2 B) e ^ {- x} \cdot \sin x + (- 2 A) e ^ {- x} \cdot \cos x y ′′ = ( 2 B ) e − x ⋅ sin x + ( − 2 A ) e − x ⋅ cos x y ′ ′ ′ = − ( 2 B ) e − x ⋅ sin x + ( 2 B ) e − x ⋅ cos x − ( − 2 A ) e − x ⋅ cos x − ( − 2 A ) e − x ⋅ sin x → y ^ {\prime \prime \prime} = - (2 B) e ^ {- x} \cdot \sin x + (2 B) e ^ {- x} \cdot \cos x - (- 2 A) e ^ {- x} \cdot \cos x - (- 2 A) e ^ {- x} \cdot \sin x \rightarrow y ′′′ = − ( 2 B ) e − x ⋅ sin x + ( 2 B ) e − x ⋅ cos x − ( − 2 A ) e − x ⋅ cos x − ( − 2 A ) e − x ⋅ sin x → y ′ ′ ′ = ( − 2 B + 2 A ) e − x ⋅ sin x + ( 2 B + 2 A ) e − x ⋅ cos x y ^ {\prime \prime \prime} = (- 2 B + 2 A) e ^ {- x} \cdot \sin x + (2 B + 2 A) e ^ {- x} \cdot \cos x y ′′′ = ( − 2 B + 2 A ) e − x ⋅ sin x + ( 2 B + 2 A ) e − x ⋅ cos x y ′ ′ ′ ′ = − ( − 2 B + 2 A ) e − x ⋅ sin x + ( − 2 B + 2 A ) e − x ⋅ cos x − ( 2 B + 2 A ) e − x ⋅ cos x − ( 2 B + 2 A ) e − x ⋅ sin x y ^ {\prime \prime \prime \prime} = - (- 2 B + 2 A) e ^ {- x} \cdot \sin x + (- 2 B + 2 A) e ^ {- x} \cdot \cos x - (2 B + 2 A) e ^ {- x} \cdot \cos x - (2 B + 2 A) e ^ {- x} \cdot \sin x y ′′′′ = − ( − 2 B + 2 A ) e − x ⋅ sin x + ( − 2 B + 2 A ) e − x ⋅ cos x − ( 2 B + 2 A ) e − x ⋅ cos x − ( 2 B + 2 A ) e − x ⋅ sin x y ′ ′ ′ ′ = ( 2 B − 2 A − 2 B − 2 A ) e − x ⋅ sin x + ( − 2 B + 2 A − 2 B − 2 A ) e − x ⋅ cos x → y ^ {\prime \prime \prime \prime} = (2 B - 2 A - 2 B - 2 A) e ^ {- x} \cdot \sin x + (- 2 B + 2 A - 2 B - 2 A) e ^ {- x} \cdot \cos x \rightarrow y ′′′′ = ( 2 B − 2 A − 2 B − 2 A ) e − x ⋅ sin x + ( − 2 B + 2 A − 2 B − 2 A ) e − x ⋅ cos x → y ′ ′ ′ ′ = ( − 4 A ) e − x ⋅ sin x + ( − 4 B ) e − x ⋅ cos x y ^ {\prime \prime \prime \prime} = (- 4 A) e ^ {- x} \cdot \sin x + (- 4 B) e ^ {- x} \cdot \cos x y ′′′′ = ( − 4 A ) e − x ⋅ sin x + ( − 4 B ) e − x ⋅ cos x
Then,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 2 e − x ⋅ sin x → y ^ {(I V)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 2 e ^ {- x} \cdot \sin x \rightarrow y ( I V ) − 2 y ′′′ + 2 y ′′ = 2 e − x ⋅ sin x → ( − 4 A ) e − x ⋅ sin x + ( − 4 B ) e − x ⋅ cos x − 2 ( ( − 2 B + 2 A ) e − x ⋅ sin x + ( 2 B + 2 A ) e − x ⋅ cos x ) + + 2 ( ( 2 B ) e − x ⋅ sin x + ( − 2 A ) e − x ⋅ cos x ) = e − x ⋅ sin x → \begin{array}{l} (- 4 A) e ^ {- x} \cdot \sin x + (- 4 B) e ^ {- x} \cdot \cos x - 2 \left((- 2 B + 2 A) e ^ {- x} \cdot \sin x + (2 B + 2 A) e ^ {- x} \cdot \cos x\right) + \\ + 2 \left((2 B) e ^ {- x} \cdot \sin x + (- 2 A) e ^ {- x} \cdot \cos x\right) = e ^ {- x} \cdot \sin x \rightarrow \\ \end{array} ( − 4 A ) e − x ⋅ sin x + ( − 4 B ) e − x ⋅ cos x − 2 ( ( − 2 B + 2 A ) e − x ⋅ sin x + ( 2 B + 2 A ) e − x ⋅ cos x ) + + 2 ( ( 2 B ) e − x ⋅ sin x + ( − 2 A ) e − x ⋅ cos x ) = e − x ⋅ sin x → ( − 4 A ) e − x ⋅ sin x + ( − 4 B ) e − x ⋅ cos x + ( 4 B − 4 A ) e − x ⋅ sin x + ( − 4 B − 4 A ) e − x ⋅ cos x + + ( 4 B ) e − x ⋅ sin x + ( − 4 A ) e − x ⋅ cos x = e − x ⋅ sin x → \begin{array}{l} (- 4 A) e ^ {- x} \cdot \sin x + (- 4 B) e ^ {- x} \cdot \cos x + (4 B - 4 A) e ^ {- x} \cdot \sin x + (- 4 B - 4 A) e ^ {- x} \cdot \cos x + \\ + (4 B) e ^ {- x} \cdot \sin x + (- 4 A) e ^ {- x} \cdot \cos x = e ^ {- x} \cdot \sin x \rightarrow \\ \end{array} ( − 4 A ) e − x ⋅ sin x + ( − 4 B ) e − x ⋅ cos x + ( 4 B − 4 A ) e − x ⋅ sin x + ( − 4 B − 4 A ) e − x ⋅ cos x + + ( 4 B ) e − x ⋅ sin x + ( − 4 A ) e − x ⋅ cos x = e − x ⋅ sin x → ( − 4 A + 4 B − 4 A + 4 B ) e − x ⋅ sin x + ( − 4 B − 4 B − 4 A − 4 A ) e − x ⋅ cos x = e − x ⋅ sin x → ( − 8 A + 8 B ) e − x ⋅ sin x + ( − 8 B − 8 A ) e − x ⋅ cos x = e − x ⋅ sin x → \begin{array}{l} (- 4 A + 4 B - 4 A + 4 B) e ^ {- x} \cdot \sin x + (- 4 B - 4 B - 4 A - 4 A) e ^ {- x} \cdot \cos x = e ^ {- x} \cdot \sin x \rightarrow \\ (- 8 A + 8 B) e ^ {- x} \cdot \sin x + (- 8 B - 8 A) e ^ {- x} \cdot \cos x = e ^ {- x} \cdot \sin x \rightarrow \\ \end{array} ( − 4 A + 4 B − 4 A + 4 B ) e − x ⋅ sin x + ( − 4 B − 4 B − 4 A − 4 A ) e − x ⋅ cos x = e − x ⋅ sin x → ( − 8 A + 8 B ) e − x ⋅ sin x + ( − 8 B − 8 A ) e − x ⋅ cos x = e − x ⋅ sin x → { − 8 A + 8 B = 1 − 8 A − 8 B = 0 → { − 8 A + 8 B = 1 − 8 A = 8 B ∣ ÷ ( − 8 ) → { − 8 ⋅ ( − B ) + 8 B = 1 A = − B → { 16 B = 1 A = − B → \left\{ \begin{array}{c} -8A + 8B = 1 \\ -8A - 8B = 0 \end{array} \right. \to \left\{ \begin{array}{c} -8A + 8B = 1 \\ -8A = 8B \mid \div (-8) \end{array} \right. \to \left\{ \begin{array}{c} -8 \cdot (-B) + 8B = 1 \\ A = -B \end{array} \right. \to \left\{ \begin{array}{c} 16B = 1 \\ A = -B \end{array} \right. \to { − 8 A + 8 B = 1 − 8 A − 8 B = 0 → { − 8 A + 8 B = 1 − 8 A = 8 B ∣ ÷ ( − 8 ) → { − 8 ⋅ ( − B ) + 8 B = 1 A = − B → { 16 B = 1 A = − B → { A = − 1 16 B = 1 16 \left\{ \begin{array}{l} A = - \frac {1}{16} \\ B = \frac {1}{16} \end{array} \right. { A = − 16 1 B = 16 1
Conclusion,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 2 e − x ⋅ sin x → y 3 ( x ) = − sin x 16 ⋅ e − x + cos x 16 ⋅ e − x y ^ {(IV)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 2 e ^ {- x} \cdot \sin x \rightarrow y _ {3} (x) = - \frac {\sin x}{1 6} \cdot e ^ {- x} + \frac {\cos x}{1 6} \cdot e ^ {- x} y ( I V ) − 2 y ′′′ + 2 y ′′ = 2 e − x ⋅ sin x → y 3 ( x ) = − 16 sin x ⋅ e − x + 16 cos x ⋅ e − x
General conclusion,
y ( I V ) − 2 y ′ ′ ′ + 2 y ′ ′ = 3 y e − x + 2 e − x ⋅ x + e − x ⋅ sin x → y ^ {(IV)} - 2 y ^ {\prime \prime \prime} + 2 y ^ {\prime \prime} = 3 y e ^ {- x} + 2 e ^ {- x} \cdot x + e ^ {- x} \cdot \sin x \rightarrow y ( I V ) − 2 y ′′′ + 2 y ′′ = 3 y e − x + 2 e − x ⋅ x + e − x ⋅ sin x → y ( x ) = y h ( x ) + y 1 ( x ) + y 2 ( x ) + y 3 ( x ) → y (x) = y _ {h} (x) + y _ {1} (x) + y _ {2} (x) + y _ {3} (x) \rightarrow y ( x ) = y h ( x ) + y 1 ( x ) + y 2 ( x ) + y 3 ( x ) → y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + 3 5 e − x + e − x ⋅ ( 2 x 5 + 28 25 ) − sin x 16 ⋅ e − x + cos x 16 ⋅ e − x → y (x) = A + B x + A _ {1} e ^ {x} \cdot \cos x + A _ {2} e ^ {x} \cdot \sin x + \frac {3}{5} e ^ {- x} + e ^ {- x} \cdot \left(\frac {2 x}{5} + \frac {2 8}{2 5}\right) - \frac {\sin x}{1 6} \cdot e ^ {- x} + \frac {\cos x}{1 6} \cdot e ^ {- x} \rightarrow y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + 5 3 e − x + e − x ⋅ ( 5 2 x + 25 28 ) − 16 sin x ⋅ e − x + 16 cos x ⋅ e − x → y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + ( 3 5 + 28 25 ) ⋅ e − x + 2 x 5 ⋅ e − x − sin x 16 ⋅ e − x + cos x 16 ⋅ e − x → y (x) = A + B x + A _ {1} e ^ {x} \cdot \cos x + A _ {2} e ^ {x} \cdot \sin x + \left(\frac {3}{5} + \frac {2 8}{2 5}\right) \cdot e ^ {- x} + \frac {2 x}{5} \cdot e ^ {- x} - \frac {\sin x}{1 6} \cdot e ^ {- x} + \frac {\cos x}{1 6} \cdot e ^ {- x} \rightarrow y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + ( 5 3 + 25 28 ) ⋅ e − x + 5 2 x ⋅ e − x − 16 sin x ⋅ e − x + 16 cos x ⋅ e − x → y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + 43 25 ⋅ e − x + 2 x 5 ⋅ e − x − sin x 16 ⋅ e − x + cos x 16 ⋅ e − x y (x) = A + B x + A _ {1} e ^ {x} \cdot \cos x + A _ {2} e ^ {x} \cdot \sin x + \frac {4 3}{2 5} \cdot e ^ {- x} + \frac {2 x}{5} \cdot e ^ {- x} - \frac {\sin x}{1 6} \cdot e ^ {- x} + \frac {\cos x}{1 6} \cdot e ^ {- x} y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + 25 43 ⋅ e − x + 5 2 x ⋅ e − x − 16 sin x ⋅ e − x + 16 cos x ⋅ e − x ANSWER
y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + 43 25 ⋅ e − x + 2 x 5 ⋅ e − x − sin x 16 ⋅ e − x + cos x 16 ⋅ e − x y (x) = A + B x + A _ {1} e ^ {x} \cdot \cos x + A _ {2} e ^ {x} \cdot \sin x + \frac {4 3}{2 5} \cdot e ^ {- x} + \frac {2 x}{5} \cdot e ^ {- x} - \frac {\sin x}{1 6} \cdot e ^ {- x} + \frac {\cos x}{1 6} \cdot e ^ {- x} y ( x ) = A + B x + A 1 e x ⋅ cos x + A 2 e x ⋅ sin x + 25 43 ⋅ e − x + 5 2 x ⋅ e − x − 16 sin x ⋅ e − x + 16 cos x ⋅ e − x
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#339914 on Dec 2023