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By taking a derivative, verify that ∫secxdx=ln(secx+tanx)+C for −2π<x<2π.
Consider the differential equation
It is given that y=0 when x=1.
Solve the differential equation using the substitution y=vx.
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Show that y=ex2∫f(x)dx is a solution to the differential equation dxdy=2xy+ex2⋅f(x).
Hence solve the differential equation dxdy=x(2y−x2ex2) given that y=0 when x=−1.
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