Find the general solution of the given differential equation. y''+y=tan, 0<t<(pi)/2

Bellenik3

Bellenik3

Open question

2022-08-19

Find the general solution of the given differential equation.
y″+y=tan⁡
0<t<π2

Answer & Explanation

Sarahi Thomas

Sarahi Thomas

Beginner2022-08-20Added 5 answers

Step 1
We were given the next equation
1) y ″+y=tan⁡t, 0<t<π2
By the theorem 4.7.2, the particular solution of the equation
y ″+p(t)y′+q(t)y=g(t)
is given by
2) Y(t)=y1(t)u1(t)+y2(t)u2(t)
where u1 is given by
3) u1(t)=−∫y2(t)g(t)W[y1, y2](t) dt 
and u2 by
4) u2(t)=∫y1(t)g(t)W[y1, y2](t) dt 
where y1 and y2 form a fundamental ser of solutions of the corresponding homogeneous equation.
Step 2
The corresponding homogeneous equation is
y ″+y=0
The roots of its charateristic equation λ2+1=0 are
λ1, 2=±i
Therefore, the homogeneous solution is
5) y=c1cos⁡t+c2sin⁡ 
Let us check if y1(t)=cos⁡  and y2(t)=sin⁡  form a fundamental set of solutions by finding their Wronskian:
W[y1, y2](t)=|y1(t)y2(t)y1'(t)y2'(t)|
=cos⁡t×cos⁡t+sin⁡t×sin⁡t=cos2t+sin2t=1
Which is always non-zero. Therefore, y1 and y2 form a fundamental set of solutions.
Step 3
Let us now find u1 and u2 using (3), (4) and g(t)=tan⁡t=sin⁡tcos⁡t
u1(t)=−∫sin⁡t×sin⁡t1×cos⁡t dt =−1−cos2tcos⁡t dt 
=−∫1cos⁡t dt +∫cos⁡t dt =−ln⁡(tan⁡t+sec⁡t)+sin⁡ 
u2(t)=∫cos⁡t×sin⁡t1×cos⁡t dt =∫sin⁡ dt =−cos⁡ 
From equation (2), we have:
Y(t)=cos⁡t×(−ln⁡(tan⁡t+sec⁡t)+sin⁡t)−sin⁡t×cos⁡ 
=−cos⁡tln⁡(tan⁡t+sec⁡t)+cos⁡tsin⁡t−sin⁡tcos⁡ 
Therefore, the particular solution of the equation (1) is Y(t)=−cos⁡tln⁡(tan⁡t+sec⁡t) 
The general solution is then given by the sum of the homogeneous and the particular solution:
y(t)=c1cos⁡t+c2sin⁡t−cos⁡tln⁡(tan⁡t+sec⁡t)

Macy Villanueva

Macy Villanueva

Beginner2022-08-21Added 3 answers

Step 1
This is non-homogeneous second order differential equation where the solution has the form y=yh+yp where yh is the solution to the associated homogeneous equation and yp is the particular solution.
y″+y=tan⁡t, o<t<π2
Step 2
First we solve the associated homogeneous equation y″+y=0 using the characteristic equation r2+1=0
r2+1=0⇒r2=−1⇒r=±i
Thus the solution to the homogeneous equation is
yh=c1cos⁡t+c2sin⁡
Step 3
Using variation of parameters, we seek a solution of the form yp=u1(t)sin⁡t+u2(t)cos⁡
yp=u1(t)sin⁡t+u2(t)cos⁡t⇒
yp′=(u1′sin⁡t+u2′cos⁡t)+(u2cos⁡t−u2sin⁡t)
Set u1′sin⁡t+u2′cos⁡t=0 then
y″p=u1′cos⁡t−u2′sin⁡t−u1sin⁡t−u2cos⁡
For yp to be a solution we must have
y″p+yp=u1′cos⁡t−u2′sin⁡t=tan⁡
Step 4
Multiply u1′sin⁡t+u2′cos⁡t=0 by sin⁡ and u1′cos⁡t−u2′sin⁡t=tan⁡ by cos⁡ and add the two resulting equations.
u1′sin⁡t+u2′cos⁡t=0⇒u1′sin2t+u2′sin⁡tcos⁡t=0
u1′cos⁡t−u2′sin⁡t=tan⁡t⇒u1′cos2t+u2′sin⁡tcos⁡t=cos⁡ttan⁡
Add and we have u1′(sin2t+cos2t)=cos⁡ttan⁡ Thus u1′=cos⁡ttan⁡t=sin⁡ and hence u1=−cos⁡
Step 5
Substitute u1′=sin⁡ in u1′sin⁡t+u2′cos⁡t=0 to solve for u2′ and hence u2
u1′=sin⁡ and u1′sin⁡t+u2′cos⁡t=0⇒
sin2t+u2′cos⁡t=0⇒u2′=−sin2tcos⁡t
=cos2t−1cos⁡t=cos⁡t−sec⁡ and hence
u2=sin⁡t−ln⁡(sec⁡t+tan⁡t)
∴yp=u1(t)sin⁡t+u2(t)cos⁡(t)
=−cos⁡tsin⁡t+[sin⁡t−ln⁡(sec⁡t+tan⁡t]cos⁡
=−cos⁡tln⁡(sec⁡t+tan⁡t)
Step 6
Recall that y=yh+yp
∴y=c1cos⁡t+c2sin⁡t−cos⁡tln⁡(sec⁡t+tan⁡t)

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