What is the solution for y^'+7=e^{7x+y-5},y(0)=5 ?

The solution for y^'+7=e^{7x+y-5},y(0)=5 is y=-ln(-x+1)-7x+5

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y^'+7=e^{7x+y-5},y(0)=5

y^{^{'}} + 7 = e^{7 x + y - 5}, y (0) = 5

y = - ln (- x + 1) - 7 x + 5

Solution steps

y^{'} (x) + 7 = e^{7 x + y - 5}

Solve using substitution:

y = - ln (- x + 1) - 7 x + 5

y = - ln (- x + 1) - 7 x + 5

n = 0 \sum \infty \frac{n + 2}{n + 1}

\frac{1}{4} x^{2} - 1, 0, 2, 8

\int_{0}^{2} \frac{1}{( x - 1 ) ^{2}} d x

y = 2 sin (x)

\frac{\partial}{\partial x} (x + y)

What is the solution for y^'+7=e^{7x+y-5},y(0)=5 ?
The solution for y^'+7=e^{7x+y-5},y(0)=5 is y=-ln(-x+1)-7x+5