What is the xcos(x)=(2y+e^{3y})y^'(x),y(0)=0 ?

The xcos(x)=(2y+e^{3y})y^'(x),y(0)=0 is y'(x)

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xcos(x)=(2y+e^{3y})y^'(x),y(0)=0

x cos (x) = (2 y + e^{3 y}) y^{^{'}} (x), y (0) = 0

y' (x)

Solution steps

Find

(2 y + e^{3 y}) y^{'} (x)

given

y (0) = 0

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

2 y + e^{3 y} : 2 \cdot 0 + e^{3 \cdot 0}

= (2 \cdot 0 + e^{3 \cdot 0}) y^{^{'}} (x)

Expand

(2 \cdot 0 + e^{3 \cdot 0}) y^{'} (x) : y^{'} (x)

= y^{'} (x)

\frac{d y}{d x} + 9 x y = 4 x, y (0) = 5

\frac{d y}{d x} y = (sin^{2} (π - x))

\frac{d P}{d t} = 0.0017 P (1 - \frac{P}{6200})

\frac{d y}{d x} + 15 x^{2} y = 2 x^{2}, y (0) = 3

\frac{d x}{d t} = 80 - \frac{x}{1000}, x = 5000, t = 0

What is the xcos(x)=(2y+e^{3y})y^'(x),y(0)=0 ?
The xcos(x)=(2y+e^{3y})y^'(x),y(0)=0 is y'(x)