Show that if ycosx = e^x then (d^2 y/dx^2) - 2 tanx dy/dx - 2y = 0

Answer 1

Let's use implicit differentiation to solve for the quantities of interest...

y * cos(x) = e^x

∂/∂x(y*cos(x)) = ∂y/∂x * cos(x) + y*(-sin(x))

∂/∂x(e^x) = e^x

∂y/∂x * cos(x) + y*(-sin(x)) = e^x

Now lets take the second derivative...

∂/∂x(∂y/∂x * cos(x) + y*(-sin(x))) = ∂²y/∂x² * cos(x) + ∂y/∂x * (-sin(x)) + ∂y/∂x*(-sin(x)) - y*cos(x)

∂/∂x(∂/∂x(e^x)) = e^x

Therefore...

∂²y/∂x² * cos(x) + ∂y/∂x * (-sin(x)) + ∂y/∂x*(-sin(x)) - y*cos(x) = e^x

∂²y/∂x² * cos(x) - 2∂y/∂x * sin(x) - y*cos(x) = e^x

Divide by cos(x) -->

∂²y/∂x² - 2∂y/∂x * sin(x)/cos(x) - y = e^x/cos(x)

∂²y/∂x² - 2∂y/∂x * tan(x) - y = y (due to definition of original function)

Finally --> ∂²y/∂x² - 2∂y/∂x * tan(x) - 2y = 0