Solve the initial-value problemdydx=cosx,y(0)=1.𝑑𝑦𝑑𝑥=cos𝑥,𝑦(0)=1.Select one:a.y=−sinx+1𝑦=−sin𝑥+1b.y=sinx+1𝑦=sin𝑥+1c.y=tanx+1𝑦=tan𝑥+1d.y=cosx+1
Question
Solution
The given differential equation is dy/dx = cos(x), with the initial condition y(0) = 1.
Step 1: Integrate both sides of the equation with respect to x.
∫dy = ∫cos(x) dx
This gives us y = sin(x) + C, where C is the constant of integration.
Step 2: Use the initial condition to find the value of C.
Substitute x = 0 and y = 1 into the equation:
1 = sin(0) + C 1 = 0 + C C = 1
So, the solution to the initial-value problem is y = sin(x) + 1.
Therefore, the correct answer is (b) y = sin(x) + 1.
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