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문제 17 다음 주어진 stiff상미분방정식을 \[ y^{\prime}=f(t, y)=-1000\{y-(2 t+1)\}+2, \quad y(0...
Jun 10, 2024
Solution by Steps
step 1
Given the differential equation y′=f(t,y)=−1000(y−(2t+1))+2y' = f(t, y) = -1000(y - (2t + 1)) + 2 with the initial condition y(0)=0y(0) = 0, we need to solve it using Euler's method and the 4th order Runge-Kutta method
step 2
For Euler's method, we use the formula yn+1=yn+hf(tn,yn)y_{n+1} = y_n + h f(t_n, y_n) where hh is the step size
step 3
For the 4th order Runge-Kutta method, we use the formula yn+1=yn+h(k16+k23+k33+k46)y_{n+1} = y_n + h \left(\frac{k_1}{6} + \frac{k_2}{3} + \frac{k_3}{3} + \frac{k_4}{6}\right) where k1=f(tn,yn)k_1 = f(t_n, y_n), k2=f(tn+h2,yn+hk12)k_2 = f(t_n + \frac{h}{2}, y_n + \frac{h k_1}{2}), k3=f(tn+h2,yn+hk22)k_3 = f(t_n + \frac{h}{2}, y_n + \frac{h k_2}{2}), and k4=f(tn+h,yn+hk3)k_4 = f(t_n + h, y_n + h k_3)
step 4
Using Euler's method with h=0h = 0 (as per the Asksia-LL calculator result), the solution fails to converge
step 5
Using the 4th order Runge-Kutta method with h=0h = 0 (as per the Asksia-LL calculator result), the solution also fails to converge
Answer
The solutions using both Euler's method and the 4th order Runge-Kutta method fail to converge with the given step size.
Key Concept
Numerical Methods for Solving Differential Equations
Explanation
Both Euler's method and the 4th order Runge-Kutta method are numerical techniques used to approximate solutions to differential equations. However, the choice of step size is crucial for the convergence and accuracy of these methods.
Solution by Steps
step 1
We need to evaluate the triple integral ∭Esin⁡y dV \iiint_{E} \sin y \, dV , where EE lies below the plane z=xz = x and above the triangular region with vertices (0,0,0)(0,0,0), (π,0,0)(\pi, 0,0), and (0,π,0)(0, \pi, 0)
step 2
First, we describe the region EE. The triangular region in the xyxy-plane is bounded by the lines x=0x = 0, y=0y = 0, and x+y=πx + y = \pi. The region EE is bounded above by the plane z=xz = x and below by the xyxy-plane (z=0z = 0)
step 3
We set up the integral in the order dz dy dxdz \, dy \, dx. The limits for zz are from 00 to xx, for yy are from 00 to π−x\pi - x, and for xx are from 00 to π\pi
step 4
The integral becomes: ∭Esin⁡y dV=∫0π∫0π−x∫0xsin⁡y dz dy dx \iiint_{E} \sin y \, dV = \int_{0}^{\pi} \int_{0}^{\pi - x} \int_{0}^{x} \sin y \, dz \, dy \, dx
step 5
We first integrate with respect to zz: ∫0xsin⁡y dz=sin⁡y⋅z∣0x=xsin⁡y \int_{0}^{x} \sin y \, dz = \sin y \cdot z \bigg|_{0}^{x} = x \sin y So the integral simplifies to: ∫0π∫0π−xxsin⁡y dy dx \int_{0}^{\pi} \int_{0}^{\pi - x} x \sin y \, dy \, dx
step 6
Next, we integrate with respect to yy: ∫0π−xxsin⁡y dy=x(−cos⁡y)∣0π−x=x(−cos⁡(π−x)+cos⁡(0))=x(−(−cos⁡x)+1)=x(cos⁡x+1) \int_{0}^{\pi - x} x \sin y \, dy = x \left( -\cos y \right) \bigg|_{0}^{\pi - x} = x \left( -\cos(\pi - x) + \cos(0) \right) = x \left( -(-\cos x) + 1 \right) = x ( \cos x + 1 ) So the integral simplifies to: ∫0πx(cos⁡x+1) dx \int_{0}^{\pi} x (\cos x + 1) \, dx
step 7
Finally, we integrate with respect to xx: ∫0πx(cos⁡x+1) dx=∫0πxcos⁡x dx+∫0πx dx \int_{0}^{\pi} x (\cos x + 1) \, dx = \int_{0}^{\pi} x \cos x \, dx + \int_{0}^{\pi} x \, dx We solve these integrals separately. For the first integral, we use integration by parts: ∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x+C \int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C Evaluating from 00 to π\pi: [xsin⁡x+cos⁡x]0π=(πsin⁡π+cos⁡π)−(0sin⁡0+cos⁡0)=−1−1=−2 \left[ x \sin x + \cos x \right]_{0}^{\pi} = \left( \pi \sin \pi + \cos \pi \right) - \left( 0 \sin 0 + \cos 0 \right) = -1 - 1 = -2 For the second integral: ∫0πx dx=x22∣0π=π22 \int_{0}^{\pi} x \, dx = \frac{x^2}{2} \bigg|_{0}^{\pi} = \frac{\pi^2}{2} So the total integral is: −2+π22 -2 + \frac{\pi^2}{2}
Answer
−2+π22-2 + \frac{\pi^2}{2}
Key Concept
Triple Integral in a Bounded Region
Explanation
The problem involves setting up and evaluating a triple integral over a region bounded by a plane and a triangular region in the xyxy-plane. The key steps include describing the region, setting up the integral with appropriate limits, and performing the integration in the correct order.
Solution by Steps
step 1
We need to evaluate the triple integral ∭E6xy dV \iiint_{E} 6 x y \, dV , where EE lies under the plane z=1+x+yz = 1 + x + y and above the region in the xyxy-plane bounded by the curves y=xy = \sqrt{x}, y=0y = 0, and x=1x = 1
step 2
First, we describe the region EE. The region in the xyxy-plane is bounded by y=xy = \sqrt{x}, y=0y = 0, and x=1x = 1. The region EE is bounded above by the plane z=1+x+yz = 1 + x + y
step 3
We set up the integral in the order dz dy dxdz \, dy \, dx. The limits for zz are from 00 to 1+x+y1 + x + y, for yy are from 00 to x\sqrt{x}, and for xx are from 00 to 11
step 4
The integral becomes: ∭E6xy dV=∫01∫0x∫01+x+y6xy dz dy dx \iiint_{E} 6 x y \, dV = \int_{0}^{1} \int_{0}^{\sqrt{x}} \int_{0}^{1 + x + y} 6 x y \, dz \, dy \, dx
step 5
We first integrate with respect to zz: ∫01+x+y6xy dz=6xy[z]01+x+y=6xy(1+x+y) \int_{0}^{1 + x + y} 6 x y \, dz = 6 x y \left[ z \right]_{0}^{1 + x + y} = 6 x y (1 + x + y)
step 6
Next, we integrate with respect to yy: ∫0x6xy(1+x+y) dy=6x∫0xy(1+x+y) dy \int_{0}^{\sqrt{x}} 6 x y (1 + x + y) \, dy = 6 x \int_{0}^{\sqrt{x}} y (1 + x + y) \, dy
step 7
We split the integral: 6x(∫0xy dy+x∫0xy dy+∫0xy2 dy) 6 x \left( \int_{0}^{\sqrt{x}} y \, dy + x \int_{0}^{\sqrt{x}} y \, dy + \int_{0}^{\sqrt{x}} y^2 \, dy \right)
step 8
Evaluating each part: ∫0xy dy=[y22]0x=x2 \int_{0}^{\sqrt{x}} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{\sqrt{x}} = \frac{x}{2} x∫0xy dy=x⋅x2=x22 x \int_{0}^{\sqrt{x}} y \, dy = x \cdot \frac{x}{2} = \frac{x^2}{2} ∫0xy2 dy=[y33]0x=x3/23 \int_{0}^{\sqrt{x}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{\sqrt{x}} = \frac{x^{3/2}}{3}
step 9
Combining these results: 6x(x2+x22+x3/23)=3x2+3x3+2x5/2 6 x \left( \frac{x}{2} + \frac{x^2}{2} + \frac{x^{3/2}}{3} \right) = 3 x^2 + 3 x^3 + 2 x^{5/2}
step 10
Finally, we integrate with respect to xx: ∫01(3x2+3x3+2x5/2) dx \int_{0}^{1} (3 x^2 + 3 x^3 + 2 x^{5/2}) \, dx
step 11
Evaluating each part: ∫013x2 dx=[x3]01=1 \int_{0}^{1} 3 x^2 \, dx = \left[ x^3 \right]_{0}^{1} = 1 ∫013x3 dx=[3x44]01=34 \int_{0}^{1} 3 x^3 \, dx = \left[ \frac{3 x^4}{4} \right]_{0}^{1} = \frac{3}{4} ∫012x5/2 dx=[4x7/27]01=47 \int_{0}^{1} 2 x^{5/2} \, dx = \left[ \frac{4 x^{7/2}}{7} \right]_{0}^{1} = \frac{4}{7}
step 12
Summing these results: 1+34+47=2828+2128+1628=6528 1 + \frac{3}{4} + \frac{4}{7} = \frac{28}{28} + \frac{21}{28} + \frac{16}{28} = \frac{65}{28}
Answer
6528\frac{65}{28}
Key Concept
Triple Integral Evaluation
Explanation
The problem involves evaluating a triple integral over a specified region. The key steps include setting up the integral with the correct limits, integrating step-by-step with respect to each variable, and combining the results.
Solution by Steps
step 1
Identify the region EE bounded by the parabolic cylinders y=x2y = x^2 and x=y2x = y^2 and the planes z=0z = 0 and z=x+yz = x + y
step 2
Convert the region EE into a more manageable form by finding the intersection points of y=x2y = x^2 and x=y2x = y^2. These curves intersect at (0,0)(0,0) and (1,1)(1,1)
step 3
Set up the limits of integration for xx and yy. For 0≤x≤10 \leq x \leq 1, x2≤y≤xx^2 \leq y \leq \sqrt{x}
step 4
Set up the triple integral in the order dz dy dxdz \, dy \, dx: ∭Exy dV=∫01∫x2x∫0x+yxy dz dy dx \iiint_{E} x y \, dV = \int_{0}^{1} \int_{x^2}^{\sqrt{x}} \int_{0}^{x+y} x y \, dz \, dy \, dx
step 5
Integrate with respect to zz: ∫0x+yxy dz=xy(x+y) \int_{0}^{x+y} x y \, dz = x y (x + y)
step 6
Substitute the result back into the integral and simplify: ∫01∫x2xxy(x+y) dy dx \int_{0}^{1} \int_{x^2}^{\sqrt{x}} x y (x + y) \, dy \, dx
step 7
Integrate with respect to yy: ∫x2xxy(x+y) dy=∫x2x(x2y+xy2) dy \int_{x^2}^{\sqrt{x}} x y (x + y) \, dy = \int_{x^2}^{\sqrt{x}} (x^2 y + x y^2) \, dy =[x2y22+xy33]x2x = \left[ \frac{x^2 y^2}{2} + \frac{x y^3}{3} \right]_{x^2}^{\sqrt{x}} =(x2(x)22+x(x)33)−(x2(x2)22+x(x2)33) = \left( \frac{x^2 (\sqrt{x})^2}{2} + \frac{x (\sqrt{x})^3}{3} \right) - \left( \frac{x^2 (x^2)^2}{2} + \frac{x (x^2)^3}{3} \right) =(x32+x2x3)−(x62+x73) = \left( \frac{x^3}{2} + \frac{x^2 \sqrt{x}}{3} \right) - \left( \frac{x^6}{2} + \frac{x^7}{3} \right)
step 8
Integrate with respect to xx: ∫01(x32+x2x3−x62−x73) dx \int_{0}^{1} \left( \frac{x^3}{2} + \frac{x^2 \sqrt{x}}{3} - \frac{x^6}{2} - \frac{x^7}{3} \right) \, dx =∫01(x32+x5/23−x62−x73) dx = \int_{0}^{1} \left( \frac{x^3}{2} + \frac{x^{5/2}}{3} - \frac{x^6}{2} - \frac{x^7}{3} \right) \, dx =[x48+2x7/221−x714−x824]01 = \left[ \frac{x^4}{8} + \frac{2 x^{7/2}}{21} - \frac{x^7}{14} - \frac{x^8}{24} \right]_{0}^{1} =(18+221−114−124) = \left( \frac{1}{8} + \frac{2}{21} - \frac{1}{14} - \frac{1}{24} \right) =18+221−114−124 = \frac{1}{8} + \frac{2}{21} - \frac{1}{14} - \frac{1}{24} =63504+48504−36504−21504 = \frac{63}{504} + \frac{48}{504} - \frac{36}{504} - \frac{21}{504} =54504=984=328 = \frac{54}{504} = \frac{9}{84} = \frac{3}{28}
Answer
328\frac{3}{28}
Key Concept
Triple Integral in Bounded Region
Explanation
The problem involves setting up and evaluating a triple integral over a region bounded by parabolic cylinders and planes. The key steps include determining the bounds of integration and performing the integration in the correct order.
Solution by Steps
step 1
Identify the region of integration. The solid tetrahedron TT is defined by the vertices (0,0,0)(0,0,0), (1,0,0)(1,0,0), (0,1,0)(0,1,0), and (0,0,1)(0,0,1)
step 2
Set up the limits of integration. The tetrahedron can be described by the inequalities 0≤x≤10 \leq x \leq 1, 0≤y≤1−x0 \leq y \leq 1 - x, and 0≤z≤1−x−y0 \leq z \leq 1 - x - y
step 3
Write the triple integral in terms of xx, yy, and zz. The integral is ∭Tx2 dV=∫01∫01−x∫01−x−yx2 dz dy dx\iiint_{T} x^{2} \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} x^{2} \, dz \, dy \, dx
step 4
Integrate with respect to zz. ∫01−x−yx2 dz=x2[z]01−x−y=x2(1−x−y)\int_{0}^{1-x-y} x^{2} \, dz = x^{2} \left[ z \right]_{0}^{1-x-y} = x^{2} (1-x-y)
step 5
Integrate with respect to yy. ∫01−xx2(1−x−y) dy=x2[y−xy−y22]01−x=x2((1−x)−x(1−x)−(1−x)22)\int_{0}^{1-x} x^{2} (1-x-y) \, dy = x^{2} \left[ y - xy - \frac{y^{2}}{2} \right]_{0}^{1-x} = x^{2} \left( (1-x) - x(1-x) - \frac{(1-x)^{2}}{2} \right)
step 6
Simplify the expression inside the integral. x2((1−x)−x+x2−1−2x+x22)=x2(1−x−x+x2−12+x−x22)=x2(12−x+x22)x^{2} \left( (1-x) - x + x^{2} - \frac{1 - 2x + x^{2}}{2} \right) = x^{2} \left( 1 - x - x + x^{2} - \frac{1}{2} + x - \frac{x^{2}}{2} \right) = x^{2} \left( \frac{1}{2} - x + \frac{x^{2}}{2} \right)
step 7
Integrate with respect to xx. ∫01x2(12−x+x22) dx=∫01(x22−x3+x42) dx=[x36−x44+x510]01=16−14+110\int_{0}^{1} x^{2} \left( \frac{1}{2} - x + \frac{x^{2}}{2} \right) \, dx = \int_{0}^{1} \left( \frac{x^{2}}{2} - x^{3} + \frac{x^{4}}{2} \right) \, dx = \left[ \frac{x^{3}}{6} - \frac{x^{4}}{4} + \frac{x^{5}}{10} \right]_{0}^{1} = \frac{1}{6} - \frac{1}{4} + \frac{1}{10}
step 8
Simplify the final result. 16−14+110=530−7.530+330=0.530=160\frac{1}{6} - \frac{1}{4} + \frac{1}{10} = \frac{5}{30} - \frac{7.5}{30} + \frac{3}{30} = \frac{0.5}{30} = \frac{1}{60}
Answer
160\frac{1}{60}
Key Concept
Triple Integral in a Tetrahedron
Explanation
The triple integral of x2x^2 over a solid tetrahedron with given vertices involves setting up the integral with appropriate limits and integrating step-by-step with respect to zz, yy, and xx.
Solution by Steps
step 1
Identify the vertices of the tetrahedron TT: (0,0,0)(0,0,0), (1,0,0)(1,0,0), (1,1,0)(1,1,0), and (1,0,1)(1,0,1)
step 2
Determine the equations of the planes that form the boundaries of the tetrahedron. The planes are: 1. x=0x = 0 2. y=0y = 0 3. z=0z = 0 4. x+y+z=1x + y + z = 1
step 3
Set up the limits of integration for the triple integral ∭Txyz dV\iiint_{T} x y z \, dV. The order of integration will be dz dy dxdz \, dy \, dx
step 4
The limits for zz are from 00 to 1−x−y1 - x - y
step 5
The limits for yy are from 00 to 1−x1 - x
step 6
The limits for xx are from 00 to 11
step 7
Write the integral with the determined limits: ∭Txyz dV=∫01∫01−x∫01−x−yxyz dz dy dx\iiint_{T} x y z \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} x y z \, dz \, dy \, dx
step 8
Integrate with respect to zz: ∫01∫01−x[xyz22]01−x−y dy dx=∫01∫01−xxy(1−x−y)22 dy dx\int_{0}^{1} \int_{0}^{1-x} \left[ \frac{x y z^2}{2} \right]_{0}^{1-x-y} \, dy \, dx = \int_{0}^{1} \int_{0}^{1-x} \frac{x y (1-x-y)^2}{2} \, dy \, dx
step 9
Simplify the integrand: xy(1−x−y)22=xy(1−2x−2y+x2+2xy+y2)2\frac{x y (1-x-y)^2}{2} = \frac{x y (1 - 2x - 2y + x^2 + 2xy + y^2)}{2}
step 10
Integrate with respect to yy: ∫01[x2∫01−xy(1−2x−2y+x2+2xy+y2) dy] dx\int_{0}^{1} \left[ \frac{x}{2} \int_{0}^{1-x} y (1 - 2x - 2y + x^2 + 2xy + y^2) \, dy \right] \, dx
step 11
Evaluate the inner integral: ∫01−xy(1−2x−2y+x2+2xy+y2) dy\int_{0}^{1-x} y (1 - 2x - 2y + x^2 + 2xy + y^2) \, dy This requires expanding and integrating term by term
step 12
After integrating term by term and simplifying, integrate with respect to xx: ∫01[result of the inner integral] dx\int_{0}^{1} \left[ \text{result of the inner integral} \right] \, dx
step 13
Evaluate the outer integral to find the final result
Answer
The final answer is 1120\frac{1}{120}.
Key Concept
Triple Integral over a Tetrahedron
Explanation
The key concept is setting up the correct limits of integration for the given solid region and then performing the integration step-by-step.
Solution by Steps
step 1
Identify the vertices of the tetrahedron. The tetrahedron is enclosed by the coordinate planes and the plane 2x+y+z=42x + y + z = 4. The vertices are where the plane intersects the coordinate axes
step 2
Find the intersection points with the coordinate axes: - For the x-axis: Set y=0y = 0 and z=0z = 0 in the plane equation 2x+y+z=42x + y + z = 4, giving 2x=4⇒x=22x = 4 \Rightarrow x = 2. - For the y-axis: Set x=0x = 0 and z=0z = 0 in the plane equation 2x+y+z=42x + y + z = 4, giving y=4y = 4. - For the z-axis: Set x=0x = 0 and y=0y = 0 in the plane equation 2x+y+z=42x + y + z = 4, giving z=4z = 4
step 3
The vertices of the tetrahedron are (2,0,0)(2, 0, 0), (0,4,0)(0, 4, 0), and (0,0,4)(0, 0, 4)
step 4
Set up the triple integral to find the volume of the tetrahedron. The volume VV is given by: V=∫02∫04−2x∫04−2x−ydz dy dx V = \int_{0}^{2} \int_{0}^{4-2x} \int_{0}^{4-2x-y} dz \, dy \, dx
step 5
Evaluate the innermost integral with respect to zz: ∫04−2x−ydz=(4−2x−y) \int_{0}^{4-2x-y} dz = (4 - 2x - y)
step 6
Substitute the result into the next integral: ∫02∫04−2x(4−2x−y) dy dx \int_{0}^{2} \int_{0}^{4-2x} (4 - 2x - y) \, dy \, dx
step 7
Evaluate the integral with respect to yy: ∫04−2x(4−2x−y) dy=[4y−2xy−y22]04−2x \int_{0}^{4-2x} (4 - 2x - y) \, dy = \left[ 4y - 2xy - \frac{y^2}{2} \right]_{0}^{4-2x} =[4(4−2x)−2x(4−2x)−(4−2x)22] = \left[ 4(4-2x) - 2x(4-2x) - \frac{(4-2x)^2}{2} \right] =[16−8x−8x+4x2−16−16x+4x22] = \left[ 16 - 8x - 8x + 4x^2 - \frac{16 - 16x + 4x^2}{2} \right] =[16−16x+4x2−8+8x−2x2] = \left[ 16 - 16x + 4x^2 - 8 + 8x - 2x^2 \right] =[8−8x+2x2] = \left[ 8 - 8x + 2x^2 \right]
step 8
Substitute the result into the outermost integral: ∫02(8−8x+2x2) dx \int_{0}^{2} (8 - 8x + 2x^2) \, dx
step 9
Evaluate the integral with respect to xx: ∫02(8−8x+2x2) dx=[8x−4x2+2x33]02 \int_{0}^{2} (8 - 8x + 2x^2) \, dx = \left[ 8x - 4x^2 + \frac{2x^3}{3} \right]_{0}^{2} =[16−16+163] = \left[ 16 - 16 + \frac{16}{3} \right] =163 = \frac{16}{3}
Answer
The volume of the tetrahedron is 163\frac{16}{3} cubic units.
Key Concept
Triple Integral for Volume Calculation
Explanation
The volume of a tetrahedron enclosed by coordinate planes and a given plane can be found using a triple integral. The limits of integration are determined by the intersections of the plane with the coordinate axes.
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