What is the (t^2+2t)(dx)/(dt)=2x+7,x(1)=1 ?

The (t^2+2t)(dx)/(dt)=2x+7,x(1)=1 is \circ t^2x(dx)/(dt)+2\circ tx(dx)/(dt)

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(t^2+2t)(dx)/(dt)=2x+7,x(1)=1

(t^{2} + 2 t) \frac{d x}{d t} = 2 x + 7, 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 + 7, 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}

\int_{1}^{5} x^{2} - 6 x + 8 d x

\int 7 x^{4} - 5 x^{3} + 5 x^{2} + C d x

\frac{7}{x}, x = 1, x = 41

x \to - 7 - lim (x^{2} + 7)

What is the (t^2+2t)(dx)/(dt)=2x+7,x(1)=1 ?
The (t^2+2t)(dx)/(dt)=2x+7,x(1)=1 is \circ t^2x(dx)/(dt)+2\circ tx(dx)/(dt)