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Given a differential equation \[ \frac{d^{2}}{d t^{2}} y(t)+11 \frac{d}{d t} y...
Dec 29, 2023
Given a differential equation d2dt2y(t)+11ddty(t)+18y(t)=0 \frac{d^{2}}{d t^{2}} y(t)+11 \frac{d}{d t} y(t)+18 y(t)=0
Solution by Steps
step 1
To solve the given second-order linear homogeneous differential equation, we look for solutions of the form y(t)=ert y(t) = e^{rt} , where r r is a constant to be determined
step 2
Substituting y(t)=ert y(t) = e^{rt} into the differential equation gives us the characteristic equation: r2ert+11rert+18ert=0. r^2 e^{rt} + 11re^{rt} + 18e^{rt} = 0.
step 3
Factoring out ert e^{rt} from the characteristic equation, we get: ert(r2+11r+18)=0. e^{rt}(r^2 + 11r + 18) = 0. Since ert e^{rt} is never zero, we set the quadratic r2+11r+18=0 r^2 + 11r + 18 = 0
step 4
Solving the quadratic equation r2+11r+18=0 r^2 + 11r + 18 = 0 by factoring, we find the roots: (r+2)(r+9)=0, (r + 2)(r + 9) = 0, which gives us r=−2 r = -2 and r=−9 r = -9
step 5
The general solution to the differential equation is a linear combination of the solutions corresponding to each root: y(t)=C1e−2t+C2e−9t, y(t) = C_1 e^{-2t} + C_2 e^{-9t}, where C1 C_1 and C2 C_2 are constants determined by initial conditions
Answer
The general solution to the differential equation is y(t)=C1e−2t+C2e−9t y(t) = C_1 e^{-2t} + C_2 e^{-9t} .
Key Concept
Solving second-order linear homogeneous differential equations using the characteristic equation.
Explanation
The characteristic equation provides the roots that determine the exponential solutions to the differential equation, which are combined to form the general solution.
The 2D Fourier transform of a 3D function U(x,y;z)U(x, y ; z) is A(fs,fy;z)A\left(f_{s}, f_{y} ; z\right). Then what would be the same 2D Fourier transform of ∂∂x∂∂yU(x,y;z)\frac{\partial}{\partial x} \frac{\partial}{\partial y} U(x, y ; z) ?
Solution by Steps
step 1
Recall the property of the Fourier transform that differentiation in the time/spatial domain corresponds to multiplication by (−i2πf)(-i 2 \pi f) in the frequency domain, where ff is the frequency variable
step 2
Apply this property to the partial derivative with respect to xx: The Fourier transform of ∂∂xU(x,y;z)\frac{\partial}{\partial x} U(x, y ; z) is (−i2πfx)A(fx,fy;z)(-i 2 \pi f_{x}) A(f_{x}, f_{y} ; z)
step 3
Apply the property again to the partial derivative with respect to yy: The Fourier transform of ∂∂y(∂∂xU(x,y;z))\frac{\partial}{\partial y} \left(\frac{\partial}{\partial x} U(x, y ; z)\right) is (−i2πfy)(-i 2 \pi f_{y}) times the Fourier transform of ∂∂xU(x,y;z)\frac{\partial}{\partial x} U(x, y ; z), which is (−i2πfx)(−i2πfy)A(fx,fy;z)(-i 2 \pi f_{x})(-i 2 \pi f_{y}) A(f_{x}, f_{y} ; z)
step 4
Simplify the expression: Multiplying the two (−i2π)(-i 2 \pi) terms together gives (−i2πfx)(−i2πfy)=−4π2fxfy(-i 2 \pi f_{x})(-i 2 \pi f_{y}) = -4 \pi^2 f_{x} f_{y}
step 5
The 2D Fourier transform of ∂∂x∂∂yU(x,y;z)\frac{\partial}{\partial x} \frac{\partial}{\partial y} U(x, y ; z) is therefore −4π2fxfyA(fx,fy;z)-4 \pi^2 f_{x} f_{y} A(f_{x}, f_{y} ; z)
Answer
The 2D Fourier transform of ∂∂x∂∂yU(x,y;z)\frac{\partial}{\partial x} \frac{\partial}{\partial y} U(x, y ; z) is −4π2fxfyA(fx,fy;z)-4 \pi^2 f_{x} f_{y} A(f_{x}, f_{y} ; z).
Key Concept
Fourier transform of derivatives
Explanation
The Fourier transform of a derivative in the spatial domain is equivalent to multiplication by (−i2πf)(-i 2 \pi f) in the frequency domain for each derivative taken, where ff is the corresponding frequency variable.
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