HELP PLEASE! Assessment Solve y' = 8zy - 2x, y(0) = 16 y(2) = ((63e^(x^2)+1)}/4
Answer all the questions in the please Solve [25 marks] 1. y' - y = 2x² cos(x), y(3) =1 [6 marks] 2. y' – 4y = 1 – e¥y', y(0) = 0 [6 marks] 3. yy' = 2y2 In(x) + 2xy2 – In(x) – - X [6 marks] 4. xy' + y = vxy, y(4) = = 1 [7 marks] End of Assessment
Use the reduction of order method to solve the following problem given one of the solution y1. (a) (x^2 - 1)y'' -2xy' +2y = 0 ,y1=x (b) (2x+1)y''-4(x+1)y'+4y=0 ,y1=e^2x (c) (x^2-2x+2)y'' - x^2 y'+x^2 y =0, y1=x (d) Prove that if 1+p+q=0 than y=e^x is a solution of y''+p(x)y'+q(x)y=0, use this fact to solve (x-1)y'' - xy' +y =0
please help to solve this differential equation. 3. Use power series solutions to solve (x+1)y"+(x-2)y' +y = 0. Center the power se- ries about the ordinary point o = 0. Write the solution as y = col first four terms..]+ ciſfirst four terms...). 4. Find the minimum radius of convergence for a power series solution to the ODE (22+2x+5)/' +10y = 0 centered about the ordinary point Xo = -6
Solve x′ =2x+y, x(0)=1 y′ =3x+4y, y(0)=0
a. With Laplace transformers solve x"+4x'+20x=0 ; x(0)=4 & x'(0)=-5 b. With Laplace transformers solve x'=2x+2y and y'=2x-y ; x(0)=1 & y(0)=2
Given: y''+2y'=2x+5-e^-2x General solution is: y=c1e^-2x+c2 +1/2(x^2)+2x+1/2(xe^-2x) Solve using the method of undetermined coefficients and show all steps please! I have the form of yp is Ax^2+Bx+Cxe^-2x, and the issue that plagues me is in solving for A B C. I get A=1/2 and I get B=2, but the terms involving C fall off the face of the earth when I substitute y' and y'' of the solution form into the equation, so how can I solve for C? Help...
(1 point) Solve the given initial value problem y′=2+e^(y−2x+4 ) y(0)=−4 The solution in the implicit form is F(x,y)=1, where F(x,y)= ? i had answer of this {y-ln(1/(-x+1))-2x+5}, don't know why its wrong.
Q1: Solve the ODE: f) Vyy' + y3/2-1. y(1) = 0. g) (2x +y)dx(2x+y-1)dy 0. i) dx=xy2e": y(2)=0. j) (1 + x*)dy + (1 + y*)dx = 0; y(1) = V3. Q1: Solve the ODE: f) Vyy' + y3/2-1. y(1) = 0. g) (2x +y)dx(2x+y-1)dy 0. i) dx=xy2e": y(2)=0. j) (1 + x*)dy + (1 + y*)dx = 0; y(1) = V3.
7c. Solve for x and y by using unimodular row reduction with initial parameters x=0 and y=1 when independent variable t=0 2x(D-2) + 6y = 0 2x + y(D-1) = 0
Systems of Equations: 3x + y = 6 2x-2y=4 Substitution: Elimination: Solve 1 equation for 1 variable. Find opposite coefficients for 1 variable. Rearrange. Multiply equation(s) by constant(s). Plug into 2nd equation Add equations together (lose 1 variable). Solve for the other variable. Solve for variable. Then plug answer back into an original equation to solve for the 2nd variable. y = 6 -- 3x solve 1" equation for y 6x +2y = 12 multiply 1" equation by 2 2x...