calculus 1 integral

Question

determine if the given function y=f(x) is a solution of the accompanying differential equation

y=1/2x integral (from 1 to 2x) of e^t / t dt

x²y' + xy = e^2x/2

Answer 1

It is not a solution, and here is why. You can tell this just by plugging in what y and y' are into the equation.

x²y' + xy

= x(d/dx(1/2x)(∫ (1 to 2x) e^t/t dt) + x *(1/2x)∫(1 to 2x) e^t/t dt

= x(d/dx(1/2x)(∫ (1 to 2x) e^t/t dt) +(1/2)∫(1 to 2x) e^t/t dt

To figure out the first term we need two tools: The product rule and the Fundamental Theorem of Calculus.

The product rule states that d/dx f(x)g(x) = f'(x)g(x)+g'(x)f(x)

The Fundamental Theorem of Calculus states that d/dx∫ (a to b) f(x)dx = f(b) - f(a).

Combining those we get for the first term:

x²*(-1/2x² ∫(1 to 2x) e^t/t dt + 1/2x*(e^2x/2x - e))

= -1/2∫(1 to 2x) e^t/t dt + x/2 (e^2x/2x - e))

Let's add this simplified version to our other term:

-1/2∫(1 to 2x) e^t/t dt + x/2 (e^2x/2x - e)) + (1/2)∫(1 to 2x) e^t/t dt

= x/2 (e^2x/2x - e) = e^2x/4 - e*x/2 ≠ e^2x/2