Dx=y, Dy=z, Dz=x Solve through systematic elimination?
Is D a variable or the differential symbol? If it's a variable: x = Dz = D (Dy) = D^2 y = D^2 (Dx) = D^3 x So, D^3 = x / x = 1 So, D = 1, omega, omega squared Assuming D to be real, D = 1 x = y = z They can have any value but the same value.
Solve the given system of differential equations by systematic elimination?
I'm not familiar with systematic elimination, but we have: y' = x y'' = x' y'' = 2x - y ... (we know x' from the other equation) y'' = 2y' - y ... (we also know x = y') y'' - 2y' + y = 0 (y' - y)' - (y' - y) = 0 Let z = y' - y. Then: z' - z = 0 z' = z z = Ae^t, for some A ... (you should know this DE well!) Thus: y' - y = z = Ae^t This is a first order linear equation, so we may use the integrating factor method. In this case, a suitable integrating factor would be e^(-t). Multiplying through: e^(-t) y' - e^(-t) y = A (e^(-t) y)' = A e^(-t) y = At + B y = Ate^t + Be^t We know: y' = x x = (Ate^t + Be^t)' = Ate^t + (A + B)e^t That's the solution!
Elimination and matrix system... I cannot figure them out on my own. It's an indepentent study course.?
Use the elimination method to find the solutions. 6. x+y+z=1 X+y-2z=3 X+2y+z=2 8. 3x-2y-z=7 Z=x+2y-5 -x+4y+2z=-4 12. x-2y+z=1 2x-3y+z=5 -x-2y+3z=-13 14. x+y+z=0 2x-5y-z=-1 3x-5y+2z=4/5 Use the matrix method to find the solutions. 30. x+3y+z=7 X+y+4z=18 -x-y+z=7 34. x+3y-2z=15 -2x-4y+5z=-32 X+4y-3z=23 36. x+2y-z=-5 3x+2y+3z=-7 5x-y-2z=-30
Solve the system of differential equations?
dx/dt= -2x+3y dy/dt= 3x-2y let y' = dy/dt and x' = dx/dt (i) x' = -2x + 3y (ii) y' = 3x - 2y either eliminate x' or y'; ~~~~~~~~~~~~~~~~~ i will eliminate x' first find the derivative y' in (ii) ie the second derivative and remember you are differentiating with respect to t and not x; so you get y'' = 3x' - 2y' from (i) x' = -2x + 3y; so replace this value of x' into y'' = 3x' - 2y' y'' = 3x' - 2y' y'' = 3(-2x + 3y) - 2y' y'' = -6x + 9y - 2y' y'' + 2y' - 9y = -6x y'' + 2y' - 9y = -2(3x) eliminate x by using equation (ii) ie y' = 3x - 2y 3x = y' + 2y y'' + 2y' - 9y = -2(3x) y'' + 2y' - 9y = -2(y' + 2y) y'' + 2y' - 9y = -2y' - 4y y'' + 4y' - 5y = 0 characteristic equation; r² + 4r - 5 = 0; r = 1, -5 y(t) = C₁e^(t) + C₂e^(-5t) initial conditions; y(0)=1 1 = C₁ + C₂ next step solve for x ~~~~~~~~~~~~~~~ use the solution of y(t) and and its derivative y'(t) to solve for x y(t) = C₁e^(t) + C₂e^(-5t) y'(t) = C₁e^(t) - 5C₂e^(-5t) replace these values in (ii) y' = 3x - 2y 3x = y' + 2y 3x = C₁e^(t) - 5C₂e^(-5t) + 2( C₁e^(t) + C₂e^(-5t) ) 3x = 3C₁e^(t) - 3C₂e^(-5t) x = C₁e^(t) - C₂e^(-5t) initial condition; x(0)=1 1 = C₁ - C₂ calculate C₁ and C₂ 1 = C₁ - C₂ 1 = C₁ + C₂ C₁ = 1 and C₂ = 0 x(t) = e^(t) y(t) = e^(t)
How do you solve [math]dy/dx + 1/x = e^y/x^2[/math]?
Let, z=e^(-y) =>dz=-e^(-y) dy =>dz=-z dynow,the given equation reduces to- -dz/dx + z/x = 1/x^2 -which is a BERNOULLI's Equation.now, multiplying both sides with integrating factor, e^(integration of -1/x),solve it.
How can I solve [math]\dfrac{dy}{dx}=x+y^2[/math]?
The given Riccati , first-order, nonlinear ordinary differential equation can be solved with the help of Mathematica by typing the following code :FullSimplify[DSolve[Derivative[1][y][x] == y[x]^2 + x, y[x], x]]The result or the solution to the given differential equation is the following :[math]\large y (x) =\displaystyle \frac { \sqrt {x}\left (c_ 1 J_ {\frac {2} {3}}\left (\frac {2 x^{3/2}} {3} \right) - J_ {-\frac {2} {3}}\left (\frac {2 x^{3/2}} {3} \right) \right)} {c_ 1 J_ {-\frac {1} {3}}\left (\frac {2 x^{3/2}} {3} \right) + J_ {\frac {1} {3}}\left (\frac {2 x^{3/2}} {3} \right)}[/math]where [math]J_n(x)[/math] is the Bessel function of the first kind.Just as an additional verification , if we take the constant [math]c_1[/math] in the solution above as equal to [math]1[/math], and if we try to find if the derivative of the Mathematica solution [math]y(x)[/math] obtained above is equal to [math]x + y(x)^2[/math] by typing :FullSimplify[D[(Sqrt[x]*(BesselJ[2/3, (2*x^(3/2))/3] - BesselJ[-(2/3), (2*x^(3/2))/3]))/(BesselJ[-(1/3), (2*x^(3/2))/3] + BesselJ[1/3, (2*x^(3/2))/3]), x] == x + ((Sqrt[x]*(BesselJ[2/3, (2*x^(3/2))/3] - BesselJ[-(2/3), (2*x^(3/2))/3]))/(BesselJ[-(1/3), (2*x^(3/2))/3] + BesselJ[1/3, (2*x^(3/2))/3]))^2]the result or answer obtained is :Truewhich verifies the solution yielded by Mathematica .As an illustrative example , below is the graph of the solution to the differential equation for the values [math]c_1 = 1 [/math], [math]c_1 = 2.5[/math] , and [math]c_1 = 5[/math] (made with Mathematica and Photoshop) . Click on the image to enlarge it :For more info see the following links :Riccati equationRiccati Differential EquationBessel Function of the First Kind
How do I solve [math]\dfrac{dy}{dx}=x^2+y^2[/math]?
I want to point out that the solution to the given differential equation with six Bessel functions can be obtained from Wolfram Alpha or with the help of Mathematica by typing the code:DSolve[y'[x] == x^2 + y[x]^2, y[x], x]This solution is:[math]\displaystyle \large y(x) = \frac{x^2 \left(c_1 J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)-c_1 J_{-\frac{5}{4}}\left(\frac{x^2}{2}\right)-2 J_{-\frac{3}{4}}\left(\frac{x^2}{2}\right)\right)-c_1 J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)}{2 x \left(c_1 J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)+J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)\right)}[/math]The solution above can be simplified by typing:FullSimplify[DSolve[y'[x] == x^2 + y[x]^2, y[x], x]]The result is a solution with four Bessel functions of the first kind:[math]\displaystyle \large \boxed{y(x) = \frac{x \left(k J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)-J_{-\frac{3}{4}}\left(\frac{x^2}{2}\right)\right)}{k J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)+J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)}}[/math]Below is a three-dimensional plot of the solution above for varying values of [math]x[/math] and [math]k[/math] (from Wolfram Alpha):As a general related result, the solution to the differential equation [math]\displaystyle a y'(x)=b x^2+c y(x)^2[/math] is found to be (verified with Mathematica):[math]\displaystyle \large y(x) = \frac{x \sqrt{\frac{b}{a}} \left(c_1 J_{\frac{3}{4}}\left(\frac{\sqrt{b c} }{2 a} x^2\right)-J_{-\frac{3}{4}}\left(\frac{\sqrt{b c} }{2 a} x^2\right)\right)}{\sqrt{\frac{c}{a}} \left(J_{\frac{1}{4}}\left(\frac{\sqrt{b c} }{2 a} x^2\right)+c_1 J_{-\frac{1}{4}}\left(\frac{\sqrt{b c} }{2 a}x^2\right) \right)}[/math]
Solve the sistem of differential equations by either systematic elimination or determinants.?
Use the substitution z = x - y dz/dt = dx/dt - dy/dt = 2x-y - x = z So dz / z = dt ln(z/C) = t z = C*e^t x - y = C*e^t Now we can return to the earlier equations dy/dt = x = y + C*e^t dy/dt - y = C*e^t The complementary solution is y = A*e^t So the particular solutions is y = B*t*e^t dy/dt = B*e^t * [t + 1] dy/dt - y = B*e^t Thus B = C Full solution of y: y = A*e^t + C*t*e^t = (A + C*t)*e^t x = y + C*e^t = (A+C+C*t)*e^t
Solve the sistem of differential equations by either systematic elimination or determinants?
I would try using Laplace transforms to solve these. The problem would translate to sX - 6Y = x0 X - sY + Z = -y0 X + Y - sZ = -z0 Now you can solve these algebraic equations for X, Y and Z in terms of s. Take the inverse Laplace transforms to get x(t), y(t) and z(t). Y(s) = A/(s-6) + B/(s+1)^2 + C/(s+1) Taking hte inverse transform y(t) = A*e^t + (B*t + C) / e^t Try the rest.