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Find the general solution of the differential equation eˣdy+(yeˣ+2x)dx=0

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Final answer:

The problem is to find the general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0, which can be solved by checking exactness or finding an integrating factor.

Step-by-step explanation:

To find the general solution of the given differential equation e^xdy + (ye^x + 2x)dx = 0, we can rearrange it and integrate.

e^xdy + (ye^x + 2x)dx = 0

e^xdy = - (ye^x + 2x)dx

Now, integrate both sides:

int e^y ,dy = - \int (ye^x + 2x) ,dx

Integrate the left side with respect to \(y\):

e^y = - int (ye^x + 2x) ,dx + C

Now, integrate the right side with respect to x:

e^y = - int ye^x ,dx - int 2x ,dx + C

e^y = -e^x(y + 2x) + C

Solve for y:

y = -2x - 1 + Ce^{-x}

So, the general solution to the given differential equation is:

y = -2x - 1 + Ce^{-x}

where C is an arbitrary constant.

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