2302-practice-mid1-soln

# Dx 2 dy dx so the equation transforms into dv dx 2 1

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dx = 2 + dy dx , so the equation transforms into dv dx 2 = 1 ve v 2 . After canceling the 2s and separating, we just need to integrate both sides: integraldisplay ve v dv dx dx = integraldisplay 1 dx, integraldisplay ve v dv = x + C, e v ( v 1) = x + C (integration by parts). Finally, we replace v by 2 x + y to obtain the solution e 2 x + y (2 x + y 1) = x + C. Because we can’t solve this equation for y ( x ), we have to be satisfied with it in implicit form.
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