y\' = e^x+y using the laws of exponential e^ x+y = e^x.e^y so, we can write the

Question

y' = e^x+y using the laws of exponential e^ x+y = e^x.e^y so, we can write the above DE as y'' = e^x.e^y This is separable and rearrange to 1/e^y dy = e^x dx or e^-y dy = e^x dx Integrating both sides we have -e^-y = e^x + C - - (1), where C = constant of integration Substituting the initial condition y(x_0) = y_0 in the above equation -e^-3n = e^3x + C or C = -e^ Substituting C in(1) Taking the natural logarithms on both sides we have -y = ln(-e^x + (e^-3n + e^3x)) General solution is y = -ln(-e^x + (e^-ya + e^xa)) Restriction for integral constant Here the integral constant is C = -e^ - e^ or e^-ya + e^xa = -C

Explanation / Answer

dy/dx = ex . ey

=> e-y . dy = ex . dx

Integrating both sides we get

-e-y = ex + c

Plugging the initial condition y( x0 ) = y0

=> -e-y0 = ex0 + c

=> c = -e-y0 - ex0

Therefore , the solution of the differential equation becomes

-e-y = ex  - e-y0 - ex0

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