`dy/dt=sint + 1`

To solve, isolate dy.

`dy=(sint +1)dt`

Then, integrate both sides.

`int dy =int(sint+1)dt`

`int dy= int sint dt + int 1dt`

To integrate, apply the formulas `int du=u +C` and `int sinudu=-cosu +C` .

`y+C=-cost +C+t +C`

Since C represents any number (constant), we may re-write the...

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`dy/dt=sint + 1`

To solve, isolate dy.

`dy=(sint +1)dt`

Then, integrate both sides.

`int dy =int(sint+1)dt`

`int dy= int sint dt + int 1dt`

To integrate, apply the formulas `int du=u +C` and `int sinudu=-cosu +C` .

`y+C=-cost +C+t +C`

Since C represents any number (constant), we may re-write the equation with one C only.

`y=-cost + t+C`

To get the value of C, use the condition y(pi/3)=1/2. So, plug-in t=pi/3 and y=1/2.

`1/2=-cos(pi/3)+pi/3+C`

`1/2=-1/2+pi/3+C`

Then, isolate C.

`1/2+1/2-pi/3=C`

`1-pi/3=C`

And, plug-in the value of C to `y=-cost +t + C` .

**Hence, the equation is:**

** `y=-cost +t + 1-pi/3`**