Ask a New Question

Question

solve the differential equation that satisfies the given condition:

dy/dx=(ycos(x))/(6+y^2), y(0)=1
9 years ago

Answers

Steve
y' = (y cosx)/(6+y^2)

(6+y^2)/y dy = cosx dx

6lny + 1/2 y^2 = sinx + c
y^2 + 12lny = 2sinx + c

I don't see any way to express y explicitly as a function of x, using elementary functions.
9 years ago

Related Questions

Solve the differential equation x^2dy+ (x-3xy + 1)=0 Solve the differential equation below (Y + y^3/3 + x^2/2)dx + 1/4(1 + y^2)xdy=0 Solve the differential equation dy/dx = √[(x^3)(y)] with initial conditions x = 1, y = 2 so f... Solve the differential equation: dy/dx = (3x^2) / (y+1) with initial conditions x = -1, y = 2 I h... solve the differential equation dy/dx=y^2/x^3 for y=f(x) with condition y(1)=1. Solve the differential equation (x+ 2)2y′′−(x+ 2)y′+y= 3x+ 4. solve the differential equation y''-4y'+4y=0 using variation of constant formula. solve the differential equation y''-3y'-4y=2sinx using method of undetermined coefficients. Solve the differential equation y"/y =(-4/18)*(1/(x^2)) Solve the differential equation. (1 - x^2)v''(x) + 2(1 + x^2)v'(x) = 0
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use