find the general solution ( y + x^(2)ye^(2x))dx - (x + xy)dy = 0 i tried to make
ID: 1889096 • Letter: F
Question
find the general solution( y + x^(2)ye^(2x))dx - (x + xy)dy = 0
i tried to make it exact but it was way too long, so I just took out y from the dx
and x from the dy and divided them both and got a separable differential equation, is this the correct method?if so can you work it out and show me what you got? would appreciate it so much thanks!
Explanation / Answer
y(1 + x^2e^(2x))dx - x(1 + y) dy = 0 (1 + x^2 e^(2x))/x dx = (1 + y)/y dy => integ 1/x + xe^(2x) dx = integ 1/y + 1 dy => ln(x) + integ xe^(2x) dx = ln y + y + c => ln(x) + x integ e^(2x) dx - integ d/dx(x) integ e^(2x) dx dx = ln y + y + c => ln(x) + xe^(2x)/2 - integ e^(2x)/2 dx = ln y + y + c => ln(x) + xe^(2x)/2 - e^(2x)/4 = ln y + y + c => ln(x) + e^(2x)/2 [ x - 1/2] = ln y + y + c
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