`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`