What is the (t^2+2t)(dx)/(dt)=2x+6,x(1)=1 ?

The (t^2+2t)(dx)/(dt)=2x+6,x(1)=1 is \circ t^2x(dx)/(dt)+2\circ tx(dx)/(dt)

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(t^2+2t)(dx)/(dt)=2x+6,x(1)=1

(t^{2} + 2 t) \frac{d x}{d t} = 2 x + 6, x (1) = 1

\circ t^{2} x \frac{d x}{d t} + 2 \circ t x \frac{d x}{d t}

Solution steps

Find

(t^{2} + 2 t) \frac{d x}{d t} \circ x (1)

given

(t^{2} + 2 t) \frac{d x}{d t} (x) = 2 x + 6, x (1) = 1

Expand

(t^{2} + 2 t) \frac{d x}{d t} \circ x \cdot 1 : \circ t^{2} x \frac{d x}{d t} + 2 \circ t x \frac{d x}{d t}

= \circ t^{2} x \frac{d x}{d t} + 2 \circ t x \frac{d x}{d t}

y^{^{''}} + 7 y = 0

\frac{d}{d x} (e^{3 x} + 3 x e^{3 x})

\frac{d}{d x} (2 x^{4 + 7 x + 5})

\int \frac{y}{( 1 + y ^{2} )} d y

\int \frac{4 x ^{2} - 3 x + 4}{x ^{3} + 4 x} d x

What is the (t^2+2t)(dx)/(dt)=2x+6,x(1)=1 ?
The (t^2+2t)(dx)/(dt)=2x+6,x(1)=1 is \circ t^2x(dx)/(dt)+2\circ tx(dx)/(dt)