What is the solution for (1+e^x)yy^'=e^x,y(0)=1 ?

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

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(1+e^x)yy^'=e^x,y(0)=1

(1 + e^{x}) y y^{^{'}} = e^{x}, y (0) = 1

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

Solution steps

(1 + e^{x}) y y^{'} (x) = e^{x}

Solve separable ODE:

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

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

f (x) = 3 x^{2} - 8 x - x

x \to - 3 lim (x^{3} - 4 x^{2} - 7)

n = 1 \sum \infty \frac{n - 9}{n + 9}

\frac{( 11 s ^{2} - 2 s + 5 )}{( s - 2 ) ( 2 s - 1 ) ( s + 1 )}

- 3 t

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