Which of the following is the solution to the differential equation ⅆyⅆx=ycosxⅆyⅆx=y⁢cos⁡x with the initial condition y(π2)=ey(π2)=e ?

Question

Which of the following is the solution to the differential equation dy/dx=ycosx with the initial condition y(pi/2)=e

y=cosx+e

y=e^(cosx+1)

y=e⋅sinx

y=e^(sinx)

Answer 1

dy/y = (cosx)dx ---------- move the y's on the left and x's on the right

Intg (dy/y)= Intg (cosx)dx --------- Take the Integral of both sides

Ln IyI = sinx +C--------------know your integral formulas esp for 1/y and trig with Cos

e^(LnIyI) = e^(sinx+C)---------take e on both sides

y = Ce^sinx ---------simplify, e Ln cancel, and the C comes infront of the e^sinx

initial position of (Pie/2,e) plug in to solve for C

e =Ce^sin(pie/2)

e=Ce -------------------SIMPLIFY, SIN(PIE/2)=1

C=1

Therefore,

Y=1e^sinx D