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Solve the initial-value problemdydx=cosx,y(0)=1.𝑑𝑦𝑑𝑥=cos⁡𝑥,𝑦(0)=1.Select one:a.y=−sinx+1𝑦=−sin⁡𝑥+1b.y=sinx+1𝑦=sin⁡𝑥+1c.y=tanx+1𝑦=tan⁡𝑥+1d.y=cosx+1

Question

Solve the initial-value problem

dydx=cos⁡x,y(0)=1 \frac{dy}{dx} = \cos{x}, \quad y(0) = 1

Select one:

  • a. y=−sin⁡x+1 y = -\sin{x} + 1
  • b. y=sin⁡x+1 y = \sin{x} + 1
  • c. y=tan⁡x+1 y = \tan{x} + 1
  • d. y=cos⁡x+1 y = \cos{x} + 1
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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.

This problem has been solved

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