Find the solution of the differential equation that satisfies the given initial condition.
y^prime tan x=a+y, y(pi / 3)=a, 0

Passo 1

We have the ODE y'tan x = a + y with y(pi/3)=a. This is a first‐order linear ODE. To put it in standard form, divide both sides by tan x (nonzero for 0) and use 1/tan x=cot x, giving
dydx - cot x,y = acot x

Passo 2

Next we find the integrating factor. For an equation of the form y' + P(x)y = Q(x), the integrating factor is mu(x)=e^integral P(x),dx. Here P(x)=-cot x, so
mu(x)=expl(!integral -cot x,dxr)=expl(-ln|sin x|r)=1sin x

Passo 3

Multiply the standard‐form ODE by the integrating factor 1/sin x. The left side becomes the derivative of y/sin x and the right side is
acot xsin x=acos xsin^2 x
So we get
ddxl( ysin xr)=acos xsin^2 x

Passo 4

Integrate both sides with respect to x. On the left we get y/sin x. On the right use the substitution u=sin x, du=cos x,dx:
integral acos xsin^2 x,dx=integral aduu^2=-au+C=-asin x+C
Thus
ysin x=-asin x+C

Passo 5

Solve for y by multiplying by sin x:
y=Csin x -a
This is the general solution.

Passo 6

Now impose the initial condition y(pi/3)=a. Since sin(pi/3)=raiz3/2, we have
a = Craiz32 - a ,Rightarrow,2a = Craiz32,Rightarrow,C=4araiz3

Passo 7

Substitute C=4a/raiz3 into the general solution to get the particular solution.

Resposta

y(x)=4araiz3sin x -a

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