how to resolve (x + 1)dx + e^y dy = 0 at points (x, y) = (0, 1).

Question

I don't understand this problem. If you can give me the answer step by step it would be perfect. It would be even more perfect if you explain it. Thanks

Answer 1

Here given differential equation is

(x+1)dx + e^ydy =0,

Separation of variables method, we have

e^ydy = -(x+1)dx

Taking Integral on both sides, we get

e^y = -[(x¹⁺¹/1+1)+x]+C

e^y = -[(x²/2)+x]+C.

Using the initial condition (x,y)=(0,1)

e =-(0²/2+0)+C,

C=e

Thus, e^y = -[(x²/2)+x] + e.

Taking 'ln' on both sides, we get

ln(e^y) = ln(-[(x²/2)+x] + e),

y ln(e) = ln(-[(x²/2)+x] + e),

y(x) = ln(-[(x²/2)+x] + e) since ln(e)=1.

Answer 2

Separate variables to get e^ydy = (-x-1)dx

Integrate both sides: e^y = -(1/2)x² - x + C

Since y = 1 when x = 0, we have C = e

So, e^y = -(1/2)x² - x + e

Therefore, y = ln (-(1/2)x² - x + e)