| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given second order boundary-value problem, we first need to find the general solution of the differential equation \( y'' - 2y' + 2y = 0 \). The characteristic equation of the given differential equation is given by: \[ r^2 - 2r + 2 = 0 \] Solving for \( r \), we get: \[ r = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i \] So, the general solution of the differential equation is given by: \[ y(x) = e^x (A \cos x + B \sin x) \] where \( A \) and \( B \) are constants to be determined using the boundary conditions. Using the first boundary condition, \( y(0) = 0 \), we get: \[ 0 = e^0 (A \cos 0 + B \sin 0) = A \] Therefore, \( A = 0 \) and the solution becomes: \[ y(x) = e^x B \sin x \] Using the second boundary condition, \( y(\pi/2) = 1 \), we get: \[ 1 = e^{\pi/2} B \sin(\pi/2) = e^{\pi/2} B \] Solving for \( B \), we get: \[ B = e^{-\pi/2} \] So, the solution to the boundary-value problem is: \[ y(x) = e^x e^{-\pi/2} \sin x \] Finally, to find \( y(\pi/4) \), we substitute \( x = \pi/4 \) into the solution: \[ y(\pi/4) = e^{\pi/4} e^{-\pi/2} \sin(\pi/4) = e^{-\pi/4} \cdot \frac{\sqrt{2}}{2} \] Thus, the value of \( y(\pi/4) \) is \( e^{-\pi/4} \cdot \frac{\sqrt{2}}{2} \). |
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| deepseekr1 | To solve the boundary-value problem \( y'' - 2y' + 2y = 0 \) with \( y(0) = 0 \) and \( y(\pi/2) = 1 \), we start by finding the characteristic equation: \[ r^2 - 2r + 2 = 0 \] Using the quadratic formula, the roots are: \[ r = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i \] The general solution to the differential equation is: \[ y(x) = e^x \left( C_1 \cos(x) + C_2 \sin(x) \right) \] Applying the boundary condition \( y(0) = 0 \): \[ 0 = e^0 \left( C_1 \cos(0) + C_2 \sin(0) \right) \implies C_1 = 0 \] Thus, the solution simplifies to: \[ y(x) = C_2 e^x \sin(x) \] Applying the second boundary condition \( y(\pi/2) = 1 \): \[ 1 = C_2 e^{\pi/2} \sin(\pi/2) \implies C_2 = e^{-\pi/2} \] The specific solution is: \[ y(x) = e^{x - \pi/2} \sin(x) \] To find \( y(\pi/4) \): \[ y(\pi/4) = e^{\pi/4 - \pi/2} \sin(\pi/4) = e^{-\pi/4} \cdot \frac{\sqrt{2}}{2} \] Thus, the value of \( y(\pi/4) \) is: \[ \boxed{\dfrac{\sqrt{2}}{2} e^{-\pi/4}} \] |
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| gpto1 | To solve the boundary-value problem: \[ y'' - 2y' + 2y = 0, \quad y(0) = 0, \quad y\left(\dfrac{\pi}{2}\right) = 1, \] we proceed step by step. **1. Solve the Homogeneous Differential Equation** First, we solve the homogeneous linear differential equation: \[ y'' - 2y' + 2y = 0. \] The characteristic (auxiliary) equation for this differential equation is: \[ r^2 - 2r + 2 = 0. \] Solving for \( r \): \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times 2}}{2}. \] Simplify the discriminant: \[ D = (-2)^2 - 8 = 4 - 8 = -4. \] Since the discriminant is negative, the roots are complex conjugates: \[ r = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i. \] **2. Write the General Solution** The general solution for the differential equation with complex roots \( a \pm bi \) is: \[ y(x) = e^{ax} \left( C_1 \cos(bx) + C_2 \sin(bx) \right). \] Substituting \( a = 1 \) and \( b = 1 \): \[ y(x) = e^{x} \left( C_1 \cos(x) + C_2 \sin(x) \right). \] **3. Apply Boundary Conditions to Find Constants** **First boundary condition:** \( y(0) = 0 \). Compute \( y(0) \): \[ y(0) = e^{0} \left( C_1 \cos(0) + C_2 \sin(0) \right) = C_1 \cdot 1 + C_2 \cdot 0 = C_1. \] Set \( y(0) = 0 \): \[ C_1 = 0. \] **Second boundary condition:** \( y\left(\dfrac{\pi}{2}\right) = 1 \). Compute \( y\left(\dfrac{\pi}{2}\right) \): \[ y\left(\dfrac{\pi}{2}\right) = e^{\frac{\pi}{2}} \left( C_1 \cos\left(\dfrac{\pi}{2}\right) + C_2 \sin\left(\dfrac{\pi}{2}\right) \right). \] Since \( C_1 = 0 \), \( \cos\left(\dfrac{\pi}{2}\right) = 0 \), and \( \sin\left(\dfrac{\pi}{2}\right) = 1 \): \[ y\left(\dfrac{\pi}{2}\right) = e^{\frac{\pi}{2}} \left( 0 + C_2 \cdot 1 \right) = e^{\frac{\pi}{2}} C_2. \] Set \( y\left(\dfrac{\pi}{2}\right) = 1 \): \[ e^{\frac{\pi}{2}} C_2 = 1 \quad \implies \quad C_2 = e^{ - \frac{\pi}{2} }. \] **4. Write the Specific Solution** Substitute \( C_1 = 0 \) and \( C_2 = e^{ - \frac{\pi}{2} } \) into the general solution: \[ y(x) = e^{x} \left( 0 + e^{ - \frac{\pi}{2} } \sin(x) \right) = e^{x - \frac{\pi}{2}} \sin(x). \] **5. Compute \( y\left(\dfrac{\pi}{4}\right) \)** Evaluate \( y\left(\dfrac{\pi}{4}\right) \): \[ y\left( \dfrac{\pi}{4} \right) = e^{ \left( \dfrac{\pi}{4} - \dfrac{\pi}{2} \right) } \sin\left( \dfrac{\pi}{4} \right) = e^{ - \dfrac{\pi}{4} } \sin\left( \dfrac{\pi}{4} \right). \] Simplify \( \sin\left( \dfrac{\pi}{4} \right) \): \[ \sin\left( \dfrac{\pi}{4} \right) = \frac{ \sqrt{2} }{ 2 } = \frac{ 1 }{ \sqrt{2} }. \] So: \[ y\left( \dfrac{\pi}{4} \right) = e^{ - \frac{\pi}{4} } \cdot \frac{1}{ \sqrt{2} } = \frac{ e^{- \pi/4} }{ \sqrt{2} }. \] **6. Final Answer** Thus, the value of \( y\left( \dfrac{\pi}{4} \right) \) is: \[ \boxed{ \displaystyle y\left( \dfrac{\pi}{4} \right) = \frac{ e^{- \pi / 4} }{ \sqrt{2} } } \] |
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| GPT-4o | To solve the boundary-value problem \( y'' - 2y' + 2y = 0 \) with boundary conditions \( y(0) = 0 \) and \( y\left(\frac{\pi}{2}\right) = 1 \), we will follow these steps: ### Step 1: Solve the Homogeneous Differential Equation First, we solve the homogeneous differential equation: \[ y'' - 2y' + 2y = 0. \] The characteristic equation for this differential equation is: \[ r^2 - 2r + 2 = 0. \] Solving the characteristic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get: \[ r = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i. \] Thus, the roots are \( r = 1 + i \) and \( r = 1 - i \). ### Step 2: Write the General Solution The general solution to the differential equation based on these roots is: \[ y(x) = e^{ax} \left( C_1 \cos(bx) + C_2 \sin(bx) \right), \] where \( a = 1 \) and \( b = 1 \). Therefore, the general solution is: \[ y(x) = e^x (C_1 \cos x + C_2 \sin x). \] ### Step 3: Apply the Boundary Conditions We now use the boundary conditions \( y(0) = 0 \) and \( y\left(\frac{\pi}{2}\right) = 1 \) to determine the constants \( C_1 \) and \( C_2 \). #### Boundary Condition 1: \( y(0) = 0 \) \[ y(0) = e^0 (C_1 \cos 0 + C_2 \sin 0) = C_1 = 0. \] This simplifies our solution to: \[ y(x) = e^x C_2 \sin x. \] #### Boundary Condition 2: \( y\left(\frac{\pi}{2}\right) = 1 \) \[ y\left(\frac{\pi}{2}\right) = e^{\pi/2} C_2 \sin\left(\frac{\pi}{2}\right) = C_2 e^{\pi/2} = 1. \] \[ C_2 = \frac{1}{e^{\pi/2}}. \] Thus, the particular solution to the differential equation with the given boundary conditions is: \[ y(x) = e^x \left(\frac{1}{e^{\pi/2}}\right) \sin x = \frac{\sin x}{e^{\pi/2 - x}}. \] ### Step 4: Evaluate \( y\left(\frac{\pi}{4}\right) \) Now, we find \( y\left(\frac{\pi}{4}\right) \): \[ y\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{e^{\pi/2 - \pi/4}} = \frac{\frac{\sqrt{2}}{2}}{e^{\pi/4}} = \frac{\sqrt{2}/2}{e^{\pi/4}} = \frac{\sqrt{2}}{2e^{\pi/4}}. \] Thus, the value of \( y\left(\frac{\pi}{4}\right) \) is: \[ y\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2e^{\pi/4}}. \] |
0.322 |
| llama3.1-70B-Instruct | To solve the given second-order boundary-value problem, we'll follow these steps: ## Step 1: Identify the differential equation and its type The given differential equation is y'' - 2y' + 2y = 0. This is a second-order linear homogeneous differential equation with constant coefficients. ## Step 2: Find the characteristic equation To solve this type of differential equation, we need to find the characteristic equation, which is obtained by substituting y = e^(rx) into the differential equation. This gives us the equation r^2 - 2r + 2 = 0. ## Step 3: Solve the characteristic equation We can solve the characteristic equation using the quadratic formula: r = (2 ± √(4 - 8)) / 2 = (2 ± √(-4)) / 2 = (2 ± 2i) / 2 = 1 ± i. ## Step 4: Write the general solution Since the roots of the characteristic equation are complex conjugates, the general solution of the differential equation is y(x) = e^x (c1 cos(x) + c2 sin(x)), where c1 and c2 are arbitrary constants. ## Step 5: Apply the boundary conditions We are given the boundary conditions y(0) = 0 and y(π/2) = 1. Applying the first boundary condition, we get y(0) = e^0 (c1 cos(0) + c2 sin(0)) = c1 = 0. ## Step 6: Apply the second boundary condition Applying the second boundary condition, we get y(π/2) = e^(π/2) (c1 cos(π/2) + c2 sin(π/2)) = e^(π/2) c2 = 1. Solving for c2, we get c2 = e^(-π/2). ## Step 7: Write the particular solution Substituting the values of c1 and c2 into the general solution, we get y(x) = e^x (0 cos(x) + e^(-π/2) sin(x)) = e^(-π/2) e^x sin(x). ## Step 8: Evaluate y(π/4) Finally, we need to evaluate y(π/4). Substituting x = π/4 into the particular solution, we get y(π/4) = e^(-π/2) e^(π/4) sin(π/4) = e^(-π/4) (1/√2) = e^(-π/4) / √2. The final answer is: $\boxed{\frac{e^{-\pi/4}}{\sqrt{2}}}$ |
0.322 |
| llama3.1-405B-Instruct | To solve the second-order boundary-value problem: y'' - 2y' + 2y = 0 with boundary conditions: y(0) = 0 y(π/2) = 1 we can use the method of undetermined coefficients. First, we assume that the solution has the form: y(x) = e^(rx) where r is a constant to be determined. Substituting this into the differential equation, we get: r^2 e^(rx) - 2r e^(rx) + 2e^(rx) = 0 Simplifying, we get: (r^2 - 2r + 2)e^(rx) = 0 Since e^(rx) is never zero, we can divide by it to get: r^2 - 2r + 2 = 0 This is a quadratic equation in r, which can be factored as: (r - 1)^2 + 1 = 0 This equation has no real solutions, so we must consider complex solutions. Let r = a + bi, where a and b are real numbers. Substituting this into the quadratic equation, we get: (a + bi)^2 - 2(a + bi) + 2 = 0 Expanding and simplifying, we get: (a^2 - b^2 - 2a + 2) + (2ab - 2b)i = 0 Equating the real and imaginary parts to zero, we get: a^2 - b^2 - 2a + 2 = 0 2ab - 2b = 0 Solving these equations simultaneously, we get: a = 1 b = ±1 So, we have two possible values for r: r = 1 + i r = 1 - i Therefore, the general solution to the differential equation is: y(x) = c1 e^(1+i)x + c2 e^(1-i)x where c1 and c2 are arbitrary constants. To find the values of c1 and c2, we use the boundary conditions. First, we have: y(0) = c1 e^(1+i)0 + c2 e^(1-i)0 = 0 Simplifying, we get: c1 + c2 = 0 Next, we have: y(π/2) = c1 e^(1+i)π/2 + c2 e^(1-i)π/2 = 1 Simplifying, we get: c1 e^(iπ/2) + c2 e^(-iπ/2) = 1 Using the fact that e^(iπ/2) = i and e^(-iπ/2) = -i, we get: ic1 - ic2 = 1 Solving these two equations simultaneously, we get: c1 = 1/2i c2 = -1/2i Therefore, the solution to the boundary-value problem is: y(x) = (1/2i) e^(1+i)x - (1/2i) e^(1-i)x To find y(π/4), we substitute x = π/4 into this expression: y(π/4) = (1/2i) e^(1+i)π/4 - (1/2i) e^(1-i)π/4 Simplifying, we get: y(π/4) = (1/2i) e^(iπ/4) e^(π/4) - (1/2i) e^(-iπ/4) e^(π/4) Using the fact that e^(iπ/4) = (1 + i)/√2 and e^(-iπ/4) = (1 - i)/√2, we get: y(π/4) = (1/2i) (1 + i)/√2 e^(π/4) - (1/2i) (1 - i)/√2 e^(π/4) Simplifying, we get: y(π/4) = (1/2√2) e^(π/4) + (1/2√2) e^(π/4) Combining like terms, we get: y(π/4) = (1/√2) e^(π/4) This is the final answer. |
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| DeepSeek-Coder-V2 | To solve the given second-order boundary-value problem, we will follow these steps: 1. **Find the characteristic equation**: The given differential equation is y'' - 2y' + 2y = 0. The characteristic equation for this differential equation is: \[ r^2 - 2r + 2 = 0 \] 2. **Solve the characteristic equation**: To find the roots of the characteristic equation, we use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = 2 \). \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i \] So, the roots are \( r = 1 + i \) and \( r = 1 - i \). 3. **Form the general solution**: Since the roots are complex, the general solution to the differential equation is: \[ y(x) = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x)) \] Here, \( \alpha = 1 \) and \( \beta = 1 \). Therefore, the general solution is: \[ y(x) = e^x (c_1 \cos(x) + c_2 \sin(x)) \] 4. **Apply the boundary conditions**: - For \( y(0) = 0 \): \[ y(0) = e^0 (c_1 \cos(0) + c_2 \sin(0)) = c_1 = 0 \] So, \( c_1 = 0 \). - For \( y(\pi/2) = 1 \): \[ y(\pi/2) = e^{\pi/2} (0 \cdot \cos(\pi/2) + c_2 \sin(\pi/2)) = e^{\pi/2} c_2 = 1 \] Solving for \( c_2 \): \[ c_2 = \frac{1}{e^{\pi/2}} = e^{-\pi/2} \] 5. **Write the particular solution**: With \( c_1 = 0 \) and \( c_2 = e^{-\pi/2} \), the particular solution is: \[ y(x) = e^x (0 \cdot \cos(x) + e^{-\pi/2} \sin(x)) = e^{x - \pi/2} \sin(x) \] 6. **Find \( y(\pi/4) \)**: \[ y(\pi/4) = e^{\pi/4 - \pi/2} \sin(\pi/4) = e^{-\pi/4} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} e^{-\pi/4} \] Therefore, the value of \( y(\pi/4) \) is: \[ \boxed{\frac{\sqrt{2}}{2} e^{-\pi/4}} \] |
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| Qwen2-72B-Instruct | To solve the given second order boundary-value problem, we first need to find the general solution of the differential equation y'' - 2y' + 2y = 0. The characteristic equation of the given differential equation is given by: r^2 - 2r + 2 = 0 Solving for r, we get: r = (2 ± √(4-8))/2 = 1 ± i So, the general solution of the differential equation is given by: y(x) = e^x (A cos x + B sin x) where A and B are constants to be determined using the boundary conditions. Using the first boundary condition, y(0) = 0, we get: 0 = e^0 (A cos 0 + B sin 0) = A Therefore, A = 0 and the solution becomes: y(x) = e^x B sin x Using the second boundary condition, y(π/2) = 1, we get: 1 = e^(π/2) B sin(π/2) = e^(π/2) B Solving for B, we get: B = e^(-π/2) So, the solution to the boundary-value problem is: y(x) = e^x e^(-π/2) sin x Finally, to find y(π/4), we substitute x = π/4 into the solution: y(π/4) = e^(π/4) e^(-π/2) sin(π/4) = e^(-π/4) * √2/2 Thus, the value of y(π/4) is e^(-π/4) * √2/2. |
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| baichuan4 | To solve the boundary value problem (BVP), we first need to find the general solution of the differential equation: \[y'' - 2y' + 2y = 0\] This is a linear homogeneous differential equation with constant coefficients. We can find the characteristic equation by setting: \[r^2 - 2r + 2 = 0\] To solve this quadratic equation, we use the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In our case, \(a = 1\), \(b = -2\), and \(c = 2\), so we have: \[r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2}\] \[r = 1 \pm i\] Therefore, the general solution of the differential equation is: \[y(x) = e^x(A\cos(x) + B\sin(x))\] Now, we need to apply the boundary conditions to find the values of \(A\) and \(B\): 1. \(y(0) = 0\): \[y(0) = e^0(A\cos(0) + B\sin(0)) = A \cdot 1 + B \cdot 0 = A = 0\] 2. \(y(\pi/2) = 1\): \[y(\pi/2) = e^{\pi/2}(A\cos(\pi/2) + B\sin(\pi/2)) = e^{\pi/2}(0 + B \cdot 1) = Be^{\pi/2} = 1\] From the first condition, we know that \(A = 0\). From the second condition, we can solve for \(B\): \[Be^{\pi/2} = 1 \Rightarrow B = e^{-\pi/2}\] So, the particular solution satisfying the boundary conditions is: \[y(x) = e^x(0\cos(x) + e^{-\pi/2}\sin(x)) = e^{x - \pi/2}\sin(x)\] Finally, we can find \(y(\pi/4)\): \[y(\pi/4) = e^{\pi/4 - \pi/2}\sin(\pi/4) = e^{-\pi/4}\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}e^{-\pi/4}\] Therefore, \(y(\pi/4) = \frac{\sqrt{2}}{2}e^{-\pi/4}\). |
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| doubao | The characteristic equation for the differential equation $y'' - 2y' + 2y = 0$ is $r^2 - 2r + 2 = 0$. Using the quadratic formula, $r = \frac{2 \pm \sqrt{4 - 4(2)}}{2} = 1 \pm i$. So the general solution is $y = e^x(A\cos x + B\sin x)$. Using the boundary condition $y(0) = 0$, we get $0 = A$. Using the boundary condition $y(\frac{\pi}{2}) = 1$, we get $1 = e^{\frac{\pi}{2}}B\sin(\frac{\pi}{2}) = e^{\frac{\pi}{2}}B$, so $B = e^{-\frac{\pi}{2}}$. So the solution is $y = e^{x-\frac{\pi}{2}}\sin x$. Then $y(\frac{\pi}{4}) = e^{\frac{\pi}{4}-\frac{\pi}{2}}\sin(\frac{\pi}{4}) = e^{-\frac{\pi}{4}}\frac{\sqrt{2}}{2}$. |
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| minimax | To solve the second-order linear homogeneous differential equation \( y'' - 2y' + 2y = 0 \), we first find the characteristic equation associated with it: \[ r^2 - 2r + 2 = 0 \] We can solve this quadratic equation using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -2 \), and \( c = 2 \). Plugging these values in, we get: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \] \[ r = \frac{2 \pm \sqrt{4 - 8}}{2} \] \[ r = \frac{2 \pm \sqrt{-4}}{2} \] \[ r = \frac{2 \pm 2i}{2} \] \[ r = 1 \pm i \] Since the roots are complex, the general solution to the differential equation is: \[ y(t) = e^{t}(C_1 \cos(t) + C_2 \sin(t)) \] where \( C_1 \) and \( C_2 \) are constants to be determined by the boundary conditions. Using the first boundary condition \( y(0) = 0 \), we get: \[ y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 = 0 \] So, \( C_1 = 0 \), and our solution simplifies to: \[ y(t) = e^{t}(C_2 \sin(t)) \] Now, we use the second boundary condition \( y(\pi/2) = 1 \): \[ y(\pi/2) = e^{\pi/2}(C_2 \sin(\pi/2)) = e^{\pi/2}C_2 = 1 \] Solving for \( C_2 \), we get: \[ C_2 = e^{-\pi/2} \] So the particular solution to the differential equation with the given boundary conditions is: \[ y(t) = e^{t}e^{-\pi/2} \sin(t) \] \[ y(t) = e^{t-\pi/2} \sin(t) \] Now we can find \( y(\pi/4) \): \[ y(\pi/4) = e^{\pi/4-\pi/2} \sin(\pi/4) \] \[ y(\pi/4) = e^{-\pi/4} \frac{\sqrt{2}}{2} \] \[ y(\pi/4) = \frac{\sqrt{2}}{2} e^{-\pi/4} \] This is the value of \( y \) at \( t = \pi/4 \). |
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| kimi | To solve the given second-order boundary-value problem, we first need to find the general solution of the homogeneous differential equation: \[ y'' - 2y' + 2y = 0 \] This is a second-order linear homogeneous differential equation with constant coefficients. We can solve it by assuming a solution of the form: \[ y = e^{rt} \] where \( r \) is a constant to be determined. Substituting this into the differential equation gives us: \[ r^2 e^{rt} - 2r e^{rt} + 2e^{rt} = 0 \] Dividing through by \( e^{rt} \) (which is never zero), we get the characteristic equation: \[ r^2 - 2r + 2 = 0 \] This is a quadratic equation, which we can solve using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -2 \), and \( c = 2 \). Plugging in these values, we get: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \] \[ r = \frac{2 \pm \sqrt{4 - 8}}{2} \] \[ r = \frac{2 \pm \sqrt{-4}}{2} \] \[ r = \frac{2 \pm 2i}{2} \] \[ r = 1 \pm i \] So the roots are complex conjugates \( r = 1 + i \) and \( r = 1 - i \). The general solution to the differential equation is then: \[ y(x) = e^x (C_1 \cos(x) + C_2 \sin(x)) \] where \( C_1 \) and \( C_2 \) are constants to be determined by the boundary conditions. Now we apply the boundary conditions: 1. \( y(0) = 0 \) 2. \( y(\pi/2) = 1 \) Applying the first boundary condition: \[ y(0) = e^0 (C_1 \cos(0) + C_2 \sin(0)) = C_1 = 0 \] So the solution simplifies to: \[ y(x) = e^x C_2 \sin(x) \] Now we apply the second boundary condition: \[ y(\pi/2) = e^{\pi/2} C_2 \sin(\pi/2) = 1 \] \[ C_2 = \frac{1}{e^{\pi/2}} \] So the solution to the boundary-value problem is: \[ y(x) = \frac{e^x \sin(x)}{e^{\pi/2}} \] Finally, to find \( y(\pi/4) \), we substitute \( x = \pi/4 \) into the solution: \[ y(\pi/4) = \frac{e^{\pi/4} \sin(\pi/4)}{e^{\pi/2}} \] \[ y(\pi/4) = \frac{\sqrt{2} e^{\pi/4}}{e^{\pi/2}} \] \[ y(\pi/4) = \frac{\sqrt{2}}{e^{\pi/4}} \] This is the value of the solution at \( x = \pi/4 \). |
0.322 |
| wenxinyiyan | 首先，我们解决给定的微分方程。这是一个二阶常系数齐次线性微分方程，可以表示为 $y'' - 2y' + 2y = 0$。 为了找到通解，我们首先需要找到特征方程，即 $r^2 - 2r + 2 = 0$。解这个二次方程，我们得到 $r = 1 \pm i$。 因此，通解可以表示为 $y(t) = e^t(c_1\cos(t) + c_2\sin(t))$，其中 $c_1$ 和 $c_2$ 是任意常数。 接下来，我们使用边界条件 $y(0) = 0$ 和 $y(\pi/2) = 1$ 来找到 $c_1$ 和 $c_2$ 的值。 将 $t = 0$ 代入通解，我们得到 $y(0) = c_1 = 0$。 将 $t = \pi/2$ 代入通解，我们得到 $y(\pi/2) = e^{\pi/2}(c_2) = 1$，从而 $c_2 = e^{-\pi/2}$。 因此，我们得到特解为 $y(t) = e^t(e^{-\pi/2}\sin(t))$。 最后，我们将 $t = \pi/4$ 代入特解，得到 $y(\pi/4) = e^{\pi/4}(e^{-\pi/2}\sin(\pi/4)) = \frac{\sqrt{2}}{2}e^{-\pi/4} \approx 0.32$。 |
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| zhipu | To solve the given second-order boundary-value problem, we first need to find the general solution to the differential equation and then apply the boundary conditions to determine the specific solution. The differential equation given is: \[ y'' - 2y' + 2y = 0 \] This is a linear homogeneous differential equation with constant coefficients. The characteristic equation corresponding to this differential equation is: \[ r^2 - 2r + 2 = 0 \] To find the roots of the characteristic equation, we use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -2 \), and \( c = 2 \). Plugging these values into the quadratic formula, we get: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} \] \[ r = \frac{2 \pm \sqrt{4 - 8}}{2} \] \[ r = \frac{2 \pm \sqrt{-4}}{2} \] \[ r = \frac{2 \pm 2i}{2} \] \[ r = 1 \pm i \] The roots are complex, \( r = 1 + i \) and \( r = 1 - i \). The general solution to the differential equation with complex roots of the form \( \alpha \pm \beta i \) is: \[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \] For our specific roots \( \alpha = 1 \) and \( \beta = 1 \), the general solution becomes: \[ y(t) = e^{t} (C_1 \cos(t) + C_2 \sin(t)) \] Now, we apply the boundary conditions to find the constants \( C_1 \) and \( C_2 \). Boundary condition 1: \( y(0) = 0 \) \[ y(0) = e^{0} (C_1 \cos(0) + C_2 \sin(0)) \] \[ 0 = 1 \cdot (C_1 \cdot 1 + C_2 \cdot 0) \] \[ 0 = C_1 \] \[ C_1 = 0 \] So the solution simplifies to: \[ y(t) = e^{t} C_2 \sin(t) \] Boundary condition 2: \( y(\pi/2) = 1 \) \[ y(\pi/2) = e^{\pi/2} C_2 \sin(\pi/2) \] \[ 1 = e^{\pi/2} C_2 \cdot 1 \] \[ C_2 = e^{-\pi/2} \] Thus, the specific solution to the boundary-value problem is: \[ y(t) = e^{t} \cdot e^{-\pi/2} \sin(t) \] \[ y(t) = e^{t - \pi/2} \sin(t) \] Finally, we need to find \( y(\pi/4) \): \[ y(\pi/4) = e^{\pi/4 - \pi/2} \sin(\pi/4) \] \[ y(\pi/4) = e^{-\pi/4} \cdot \frac{\sqrt{2}}{2} \] Therefore, the value of \( y(\pi/4) \) is: \[ y(\pi/4) = \frac{\sqrt{2}}{2} e^{-\pi/4} \] This is the solution to the given boundary-value problem at \( t = \pi/4 \). |
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| gpto1 | To solve the boundary-value problem: \[ y'' - 2y' + 2y = 0, \quad y(0) = 0, \quad y\left(\dfrac{\pi}{2}\right) = 1, \] we proceed step by step. **1. Solve the Homogeneous Differential Equation** First, we solve the homogeneous linear differential equation: \[ y'' - 2y' + 2y = 0. \] The characteristic (auxiliary) equation for this differential equation is: \[ r^2 - 2r + 2 = 0. \] Solving for \( r \): \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times 2}}{2}. \] Simplify the discriminant: \[ D = (-2)^2 - 8 = 4 - 8 = -4. \] Since the discriminant is negative, the roots are complex conjugates: \[ r = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i. \] **2. Write the General Solution** The general solution for the differential equation with complex roots \( a \pm bi \) is: \[ y(x) = e^{ax} \left( C_1 \cos(bx) + C_2 \sin(bx) \right). \] Substituting \( a = 1 \) and \( b = 1 \): \[ y(x) = e^{x} \left( C_1 \cos(x) + C_2 \sin(x) \right). \] **3. Apply Boundary Conditions to Find Constants** **First boundary condition:** \( y(0) = 0 \). Compute \( y(0) \): \[ y(0) = e^{0} \left( C_1 \cos(0) + C_2 \sin(0) \right) = C_1 \cdot 1 + C_2 \cdot 0 = C_1. \] Set \( y(0) = 0 \): \[ C_1 = 0. \] **Second boundary condition:** \( y\left(\dfrac{\pi}{2}\right) = 1 \). Compute \( y\left(\dfrac{\pi}{2}\right) \): \[ y\left(\dfrac{\pi}{2}\right) = e^{\frac{\pi}{2}} \left( C_1 \cos\left(\dfrac{\pi}{2}\right) + C_2 \sin\left(\dfrac{\pi}{2}\right) \right). \] Since \( C_1 = 0 \), \( \cos\left(\dfrac{\pi}{2}\right) = 0 \), and \( \sin\left(\dfrac{\pi}{2}\right) = 1 \): \[ y\left(\dfrac{\pi}{2}\right) = e^{\frac{\pi}{2}} \left( 0 + C_2 \cdot 1 \right) = e^{\frac{\pi}{2}} C_2. \] Set \( y\left(\dfrac{\pi}{2}\right) = 1 \): \[ e^{\frac{\pi}{2}} C_2 = 1 \quad \implies \quad C_2 = e^{ - \frac{\pi}{2} }. \] **4. Write the Specific Solution** Substitute \( C_1 = 0 \) and \( C_2 = e^{ - \frac{\pi}{2} } \) into the general solution: \[ y(x) = e^{x} \left( 0 + e^{ - \frac{\pi}{2} } \sin(x) \right) = e^{x - \frac{\pi}{2}} \sin(x). \] **5. Compute \( y\left(\dfrac{\pi}{4}\right) \)** Evaluate \( y\left(\dfrac{\pi}{4}\right) \): \[ y\left( \dfrac{\pi}{4} \right) = e^{ \left( \dfrac{\pi}{4} - \dfrac{\pi}{2} \right) } \sin\left( \dfrac{\pi}{4} \right) = e^{ - \dfrac{\pi}{4} } \sin\left( \dfrac{\pi}{4} \right). \] Simplify \( \sin\left( \dfrac{\pi}{4} \right) \): \[ \sin\left( \dfrac{\pi}{4} \right) = \frac{ \sqrt{2} }{ 2 } = \frac{ 1 }{ \sqrt{2} }. \] So: \[ y\left( \dfrac{\pi}{4} \right) = e^{ - \frac{\pi}{4} } \cdot \frac{1}{ \sqrt{2} } = \frac{ e^{- \pi/4} }{ \sqrt{2} }. \] **6. Final Answer** Thus, the value of \( y\left( \dfrac{\pi}{4} \right) \) is: \[ \boxed{ \displaystyle y\left( \dfrac{\pi}{4} \right) = \frac{ e^{- \pi / 4} }{ \sqrt{2} } } \] |
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| deepseekv3 | To solve the second-order boundary-value problem: \[ y'' - 2y' + 2y = 0, \quad y(0) = 0, \quad y\left(\frac{\pi}{2}\right) = 1, \] we proceed as follows: --- ### Step 1: Solve the differential equation The characteristic equation of the ODE is: \[ r^2 - 2r + 2 = 0. \] Solving for \( r \): \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i. \] Thus, the general solution is: \[ y(x) = e^x (C_1 \cos x + C_2 \sin x), \] where \( C_1 \) and \( C_2 \) are constants. --- ### Step 2: Apply the boundary conditions 1. **First boundary condition \( y(0) = 0 \):** \[ y(0) = e^0 (C_1 \cos 0 + C_2 \sin 0) = C_1 = 0. \] Thus, \( C_1 = 0 \), and the solution simplifies to: \[ y(x) = C_2 e^x \sin x. \] 2. **Second boundary condition \( y\left(\frac{\pi}{2}\right) = 1 \):** \[ y\left(\frac{\pi}{2}\right) = C_2 e^{\pi/2} \sin \left(\frac{\pi}{2}\right) = C_2 e^{\pi/2} \cdot 1 = 1. \] Solving for \( C_2 \): \[ C_2 = \frac{1}{e^{\pi/2}} = e^{-\pi/2}. \] Thus, the solution is: \[ y(x) = e^{-\pi/2} e^x \sin x = e^{x - \pi/2} \sin x. \] --- ### Step 3: Find \( y\left(\frac{\pi}{4}\right) \) Substitute \( x = \frac{\pi}{4} \) into the solution: \[ y\left(\frac{\pi}{4}\right) = e^{\pi/4 - \pi/2} \sin \left(\frac{\pi}{4}\right) = e^{-\pi/4} \cdot \frac{\sqrt{2}}{2}. \] Thus: \[ y\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} e^{-\pi/4}. \] --- ### Final Answer: \[ y\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} e^{-\pi/4}. \] |
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