Question

Let y'(x) + y(x)g'(x) = g(x)g'(x), y(0) = 0, x ∈ R, where f'(x) denotes df(x)/dx and g(x) is a given non-constant differentiable function on R with g(0) = g(2) = 0. The value of y(2) is ...

Answer 1

The given differential equation is

y'(x) + y(x)g'(x) = g(x)g'(x)

(dy/dx) + g'(x)y = g(x)g'(x)

which is a linear differential equation.

Thus. IF = e^{∫g'(x) dx} = e^g(x)

Therefore, the solution is

y.e^g(x) = ∫(g(x) ∫g'(x)) e^g(x) dx

y.e^g(x) = e^g(x) (g(x) – 1) + c

When x = 0, y = 0, then c = 1

Thus, the curve is 

y.e^g(x) = e^g(x) (g(x) – 1) + 1

When x = 2, then y = (0 – 1) + 1 = 0.