Answer to Question #226266 in Differential Equations for mohammed

Question

If yp = uj (t) cos(2t) + u2(t) sin(2t) is a particular solution of the differential equation

y" + 4y= csc(2t) cot(2t), then u, (t) =

Select one:

O none of the given answers is true

O Lt + - cot(2t)

O _ In | sin(2t) |

O

_ In | sin(2t) |

O

cot (2t)

Answer 1

$y'' + 4y= csc(2t) cot(2t)$

First, let's find the complementary solution of the ODE.

The characteristic equation is $r^2+4=0\Rightarrow r=\pm2i \\y_c(t)=Acos 2t+Bsin2t$

To find a particular integral of the ODE, we calculate the Wronskian:

with: $y_1(t) = cos2t\ \text{and} \ y_2(t) = sin2t \\ W(y_1, y_2) = y_1(t)y'_2 (t) − y_2(t)y'_1 (t)$

$=cos2t.(2cos2t)-sin2t(-2sin2t)\\=2(cos^2t+sin^2t)\\=2$

Then a particular integral of the ODE is:

$y_{p}(t)=-y_{1}(t) \int \frac{y_{2}(t) f(t)}{W\left(y_{1}, y_{2}\right)} d t+y_{2}(t) \int \frac{y_{1}(t) f(t)}{W\left(y_{1}, y_{2}\right)} d t \\$

Here, $f(t)=csc(2t)cot(2t)$

$y_{p}(t)=-cos2t \int \frac{sin2t. csc(2t)cot(2t)}{2} d t+sin2t \int \frac{cos2t.csc(2t)cot(2t)}{2} d t \\=-cos2t \int \frac{cot(2t)}{2} d t+sin2t \int \frac{cot^2(2t)}{2} d t \\=-\frac{cos2t.ln|sin 2t|}{4}+ sin2t \int \frac{csc^2(2t)-1}{2} d t \\=-\frac{cos2t.ln|sin 2t|}{4}+ \frac{sin2t(-cot(2t)-2t)}{4}$

Comparing the above result with the given, we will get

$u_1(t)=-\frac{ln|sin 2t|}{4}, u_2(t)= \frac{(-cot(2t)-2t)}{4}$