Find the particular solution of the differential equation

Question

Find the particular solution of the differential equation

𝑑𝑦/𝑑𝑥+𝑦cos(𝑥)=4cos(𝑥)

satisfying the initial condition y(0)=6

Your answer should be a function of x.

Answer 1

This can be rewritten as a first order separable ODE. It is in the form dy/dx × N(y) = M(x)

Solving the equation by rearranging in this form makes it out to be:

dy/dx × (1⁄_y-4) = -cos(x)

Where N(y) = (1⁄_y-4) and M(x) = -cos(x)

Integrating the equation on both sides you get:

∫(1⁄_y-4) dy = - ∫cos(x) dx

ln(y-4) = -sin(x) + C , where C∈ ΙR

C, our constant of integration, belongs to a real number

To get rid of the natural log, we must do the inverse. That means raising everything to the power of e , or Euler's number.

e^ln(y-4) = e^{-sin(x) + C}

Then recall: e^ln(x) = x

e^ln(y-4) = y-4

This simplifies to:

y-4 = e^{-sin(x) + C}

Isolating for y by adding 4 to the other side:

y = e^{-sin(x) + C} + 4

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Now to find the particular solution, we must solve for our constant of integration, C

Since y(0) = 6

x = 0 , y = 6

y(0) = 6 = e^{-sin(0) + C} + 4

6 = e^{-sin(0) + C} + 4

6 = e^{0 + C} + 4

6 = e^C + 4

6 − 4 = e^C

2 = e^C

ln(2) = ln(e^C)

ln(2) = C

Plugging C back in for our equation for y our particular solution is:

y_p(x) = e^{-sin(x) + ln(2)} + 4

Answer 2

The integrating factor is

exp(int cosx dx) = exp(sinx).

So, a solution is given by

exp(-sinx)*int exp(sinx)*4cosx dx = exp(-sinx)*4*exp(sinx) = 4.

Consider now the homoegeneous problem:

y'+cosx*y = 0 and use separation of variables.

y' = -cosx *y

dy/y = -cosx dx

ln|y| = -sinx+C

y = Ce^(-sinx)

So, the general solution is 4+Ce^(-sinx). Applying the initial condition y(0) = 6 gives

6 = 4+C, so that C = 2.

The solution is:

y(x) = 4+2exp(-sinx)