solve the IVP 1.) (e^x+y)dx+(2+x+ye^y)dy=0 y(0)=1 Solution 1) Change it to as: x
ID: 2976166 • Letter: S
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solve the IVP 1.) (e^x+y)dx+(2+x+ye^y)dy=0 y(0)=1Explanation / Answer
1) Change it to as: xy' + y = e^x 2) Integrate both sides, ==> Integral of [xy' + y] = Integral of (e^x) 3) ==> Integral of [d(xy)/dx] = Integral of (e^x) 4) ==> xy = e^x + c 5) The above implies "y = [e^x + c]/(x)" 6) The expression in (5), does not have domain (0); hence it cannot be evaluated for x = 0; As such you may verify y(0) = 1. This is an exact differential equation, and can be rewritten this way: x dy/dx = -y + e^x x dy = (-y + e^x) dx (y - e^x) dx + x dy = 0 Solution to exact differential equation is f(x,y) = C, where ?f/?x = y - e^x . . . partial derivative of f(x,y) with respect to x ?f/?y = x . . . . . . . partial derivative of f(x,y) with respect to y We can tell this is exact, since ?/?y (y - e^x) = 1 = ?/?x (x) Now we can find f(x,y) by integrating (y - e^x) dx and x dy f(x,y) = ? (y - e^x) dx = xy - e^x + g(y) f(x,y) = ? x dy = xy + h(x) Comparing both values of f(x,y), we find that g(y) = 0 and h(x) = -e^x Therefore f(x,y) = xy - e^x and f(x,y) = C xy - e^x = C Using initial condition y(0) = 1, we get 0*1 - e^0 = C 0 - 1 = C C = -1 xy - e^x = -1 xy - e^x + 1 = 0 -------------------- To answer your question: How is xdy/dx +y = d(xy)/dx d(xy)/dx = d/dx (xy) . . . . . . . . use product rule . . . . . . = x dy/dx + dx/dx * y . . . . . . = x dy/dx + y
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