What is the solution for y^'-x*e^y=0,y(0)=2 ?

The solution for y^'-x*e^y=0,y(0)=2 is y=-ln(-(x^2)/2+1/(e^2))

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y^'-x*e^y=0,y(0)=2

y^{^{'}} - x \cdot e^{y} = 0, y (0) = 2

y = - ln (- \frac{x ^{2}}{2} + \frac{1}{e ^{2}})

Solution steps

y^{'} (x) - x \cdot e^{y} = 0

Solve separable ODE:

y = - ln (- \frac{x ^{2}}{2} + \frac{1}{e ^{2}})

y = - ln (- \frac{x ^{2}}{2} + \frac{1}{e ^{2}})

\frac{d p}{d t} = 4 pt, p (1) = 7

y^{^{'}} = (sin (x)) (cos^{2} (x)) (tan (y))

\frac{d y}{d x} = y (- 8 - y)

(y^{2} + 2) d x + 3 \cdot x \cdot y d y = 0

\frac{d Q}{d t} = 2 c - Q

What is the solution for y^'-x*e^y=0,y(0)=2 ?
The solution for y^'-x*e^y=0,y(0)=2 is y=-ln(-(x^2)/2+1/(e^2))