What is the solution for y^'=e^{x-y},y(0)=ln(7) ?

The solution for y^'=e^{x-y},y(0)=ln(7) is y=ln(e^x+6)

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y^'=e^{x-y},y(0)=ln(7)

y^{^{'}} = e^{x - y}, y (0) = ln (7)

y = ln (e^{x} + 6)

Solution steps

y^{'} (x) = e^{x - y}

Solve separable ODE:

y = ln (e^{x} + 6)

y = ln (e^{x} + 6)

\frac{d y}{d x} + x^{2} y = 5 x^{2}

\frac{d x}{d y} = k x (y)

y^{^{'}} = 2 (t + 1) y - 1

\frac{d v}{d t} = 32.2 - \frac{1}{40} v

\frac{d P}{d t} = (0.1) P - (0.05) P (t)^{2}

What is the solution for y^'=e^{x-y},y(0)=ln(7) ?
The solution for y^'=e^{x-y},y(0)=ln(7) is y=ln(e^x+6)