in the equation above which of the following is a possible value of x+1


What is the panicular solution of a differential equation?

    f(x) be the panicular solution to the given differential equation with initial condition f(l) = 0. Write an equation for the line tangent to the graph of y = f (x) at x = 1.

Which function satisfies the differential equation and the given initial condition?

    So, the function y(x) that satis?es the differential equation and the given initial condition is: y(x) = 11e?x+x? 1. 5 1.1.45SupposeapopulationP ofrodentssatis?esthedifferentialequation dP/dt = kP2.

What is the value of Y (X) that represents the differential equation?

    Solving for C we get: 10 = y(0) = Ce?0+0?1, ? C = 11. So, the function y(x) that satis?es the differential equation and the given initial condition is: y(x) = 11e?x+x? 1. 5

What is x(t) if x(0) = 0?

    If x(0) = 0this implies C = 0, so x(t) = a(2t? sin(st)). On the other hand, we have y(t) = 2asin2t = 2a  1?cos(2t) 2  = a(1? cos(2t)). So, combining what we’ve derived we get: x(t) = a(2t?sin(2t)) y(t) = a(1? cos(2t)).
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