Find dy/dx and d²y/dx². x = e^t, y = te^-t

Solution:

x = e^t

dx/dt = de^t/dt = e^t

dy/dt = d( te^-t)/dt

dy/dt = tde^-t/dt + e^-td(t)/dt

dy/dt = -te^-t + e^-t = (1 - t)e^-t

dy/dx = dy/dt/dx/dt = (1 - t)e^-t /e^t

dy/dx = (1 - t)e^-2t ---(1)

d²y/dx² = d/dx(dy/dx)

Since x = e^t and y = te^-t

xy = e^t × te^-t

xy = t

Hence we can write equation (1) as

dy/dx = (1 - xy) × e^-2t

We know

e^-2t = 1/e^2t = (1/e^t ) × (1/e^t ) = (1/x) × (1/x) = 1/x²

dy/dx = (1 - xy) × (1/x²)

dy/dx = 1/x² - y/x

Since d²y/dx² = d/dx(dy/dx) we can write

d(1/x² - y/x)/dx = d(1/x²)/dx - d(y/x)/dx

(-2)x⁻³ - [ xdy/dx - ydx/dx]/x²

d²y/dx² = -2/x³ - (x/x²)dy/dx - (y/x²)

We know, dy/dx = 1/x² - y/x

d²y/dx² = -2/x³ - (1/x)(1/x - y/x) - y/x²

d²y/dx² = -2/x³ - 1/x² + y/x² - y/x²

d²y/dx² = 1/x²(-1 - 2/x)

d²y/dx² = -1/x²(1 + 2/x)

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Find dy/dx and d²y/dx². x = e^t, y = te^-t

Summary:

dy/dx = dy/dx = 1/x² - y/x and d²y/dx² = -1/x²(1 + 2/x) when x = e^t, y = te^-t