Find a particular solution of the differential equation: (dy/dx) + y(cos(x)) = 2cos(x)?
This equation is separable. Subtract the second terms from the left-hand side and factor out cos (x): dy/dx = (2-y) cos(x), so (1/(2-y)) dy = cos(x) dx. Integrate: ln(2-y) = -sin(x) + c1 Exponentiate (use e^(both sides)): 2 - y = e^(-sin(x) + c1) 2 - y = e^(-sin(x)) e^(c1) 2 - y = c2*e^(-sin(x)) (where c2=e^(c1)) y - 2 = -c2*e^(-sin(x)) y = 2 - c2*e^(-sin(x)) (general solution) OK. Since y(0) = 4, substitute in x = 0, y = 4: 4 = 2 - c2*e^(-sin(0)) 4 = 2 - (c2 * 1) so c2 = -2, and the particular solution is y = 2 + 2*e^(-sin(x)). Check it to verify that it indeed works!
Solve the separable differential equation dy/dt= 9y^4 and find the particular solution satisfying the initial?
dy/(9y^4) = dt -1/(27y^3) = t + C 27y^3 = - 1/(t + C) y^3 = - 1/(27(t + C)) y = - 1/(3 ³√(t + C) ) y(0) = - 3 -3 = - 1/(3 ³√(0 + C) ) ³√C = 1/9 C = 1/729 y = - 1/(3 ³√(t + 1/729) ) y = - 3/³√(729 t + 1)
Solve the separable differential equation and find the particular solution satisfying the initial condition ?
Let's check your general solution. You have: dy/dx = sqrt(17 - 2y) Separate the variables; dy/(17 - 2y) = dx -sqrt(17 - 2y) = x + c where c is the constant of integration. sqrt(17 - 2y) = -(x + c) 17 - 2y = (x + c)^2 2y = 17 - (x + c)^2 y(x) = [17 - (x + c)^2]/2 which is the same answer you had. Now use the initial condition to solve for c, but it will be simpler to do so if we go back to the form of the solution that looks like: -sqrt(17 - 2y) = x + c We know that y(-3) = 4, so: -sqrt(17 - 8) = c - 3 sqrt(9) = 3 - c c = 0 So the particular solution is y(x) = [17 - x^2]/2
Solve the separable differential equation dy/dt?
dy/dt= 9y^4 , y(0)= -3 ∫ dy/y^4 = 9∫ dt ∫ y^(-4)dy = 9 ∫ dt -1/3y^(-3) = 9t + C when y(0) = - 3 - 1/[3(-3)^3] = 9(0) + C 1/81 = C -1/3y^(-3) = 9t + 1/81 .......................729t + 1 [- 1/[3y^3] =------------------]81y^3 ........................81 - 27 = y^3[729t + 1] ............- 27 y^3 = ---------------- ...........729t + 1 ..............- 27 t = ∛[----------------] ............729t + 1 ..............- 1...............................3.∛(- 1). t = 3∛(--------------) or ..- --------------------] answer// .............729 + 1.....................∛(-279t- 1) dy/dx= (x-3)e^(-2y) , y(3)=ln(3) ∫ dy/e^(-2y) =∫ (x - 3)dx ∫ e^(2y)dy = ∫ (x - 3)dx 1/2e^(2y) = 1/2x^2 - 3x + C when y(3) = ln(3) 1/2e^[2ln(3)] = 1/2(3)^2 - 3(3) + C 1/2e^[ln(3)^2] = - 9/2 + C 1/2e^[ln(9)] = - 9/2 + C 1/2(9) = - 9/2 + C 9/2 + 9/2 = C C = 9 [1/2e^(2y) = 1/2x^2 - 3x + 9]2 e^(2y) = x^2 - 6x + 18 2y = ln(x^2 - 6x + 18) y = 1/2ln(x^2 - 6x + 18) answer//
Solve the separable differential equation dy/dx=6y and find the particular solution satisfying the initial condition y(0)=-4 y(x)=?
dy/dx = 6y 1/y dy = 6 dx ln y = 6x + c y = C e^6x y(0) = -4 C = -4 y = -4 e^6x
Solve the separable differential equation?
Separation of variables: dy/y^4 = -9 dt Integrate both sides: ∫ dy/y^4 = ∫ -9 dt -1/(3y³) = -9t + c If y(0) = -5... -1/(3(-5)³) = -9(0) + c 1/375 = c Therefore.. -1/(3y³) = -9t + 1/375 -3y³ = -1/(-9t + 1/375) -3y³ = -375/(-1125t + 1) y = ∛(125/(-1125t + 1)) I hope this helps!
What is the solution of the differential equation (dy/dx) ^2-xdy/dx+y=0?
(dy/dx)^2 -x(dy/dx)+y=0representing dy/dx by D we have:D^2 - xD+y=0Or, y=xD-D^2Or, dy/dx=xdD/dx +D-2DdD/dxOr, 0=xdD/dx-2DdD/dxdD/dx=0;x-2D=0So, x=2dy/dxInteg xdx= Integ 2 dyOr, x^2 /2 =2yOr, y=x^2/4+ k
Differential Equations Help Please?
These links have step by step solutions to your problems: 1. Solve the separable differential equation and find the particular solution satisfying the initial condition y(0) = 5 dy/dt = 9y^4 https://www.symbolab.com/solver/ordinary... 2. Find the solution of the differential equation which satisfies the initial condition y(0)=1. 3e^(2x) dy/dx = -4(x/y^2) https://www.symbolab.com/solver/ordinary... Hope this helps