What is the solution for 12yy^'-7e^x=0 ?

The solution for 12yy^'-7e^x=0 is y=sqrt((7e^x+c_{1))/6},y=-sqrt((7e^x+c_{1))/6}

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12yy^'-7e^x=0

12 y y^{^{'}} - 7 e^{x} = 0

y = \frac{7 e ^{x} + c _{1}}{6}, y = - \frac{7 e ^{x} + c _{1}}{6}

Solution steps

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

Solve separable ODE:

y = \frac{7 e ^{x} + c _{1}}{6}, y = - \frac{7 e ^{x} + c _{1}}{6}

y = \frac{7 e ^{x} + c _{1}}{6}, y = - \frac{7 e ^{x} + c _{1}}{6}

\frac{\partial}{\partial x} ((sin (x))^{2})

\int \frac{1}{( x + 4 ) ( x - 1 )} d x

y = e^{x}, y = e^{- 3 x}, x = ln (5)

\int x^{2} + 2 x + 5 d x

f (x) = x^{2} + 8, (4, 24)

What is the solution for 12yy^'-7e^x=0 ?
The solution for 12yy^'-7e^x=0 is y=sqrt((7e^x+c_{1))/6},y=-sqrt((7e^x+c_{1))/6}