What is the solution for 3ydy=(2sin(2x)-e^x)ydx ?

The solution for 3ydy=(2sin(2x)-e^x)ydx is y= 1/3 (-cos(2x)-e^x)+c_{1}

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3ydy=(2sin(2x)-e^x)ydx

3 y d y = (2 sin (2 x) - e^{x}) y d x

y = \frac{1}{3} (- cos (2 x) - e^{x}) + c_{1}

Solution steps

3 y d y = (2 sin (2 x) - e^{x}) y d x

Let

y

be the dependent variable. Divide by

d x

3 y \frac{d y}{d x} = (2 sin (2 x) - e^{x}) y

Substitute

\frac{d y}{d x}

with

y^{'} (x)

3 y y^{^{'}} (x) = (2 sin (2 x) - e^{x}) y

Isolate

y^{'} (x) : y^{'} (x) = \frac{2 sin ( 2 x ) - e ^{x}}{3}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{2 sin ( 2 x ) - e ^{x}}{3} d x

\int \frac{2 sin ( 2 x ) - e ^{x}}{3} d x = \frac{1}{3} (- cos (2 x) - e^{x}) + c_{1}

y = \frac{1}{3} (- cos (2 x) - e^{x}) + c_{1}

\frac{\partial}{\partial x} (3 x^{6} y^{2})

\frac{d}{d x} (- 5 (2 x - 100)^{6})

\frac{d}{d t} (6 t^{2} - 8 + 40 cos (π t))

\int 3 t^{4} d t

\frac{\partial}{\partial s} (t cos (s))

What is the solution for 3ydy=(2sin(2x)-e^x)ydx ?
The solution for 3ydy=(2sin(2x)-e^x)ydx is y= 1/3 (-cos(2x)-e^x)+c_{1}