Differential equations \(\displaystyle{\left({x}+{1}\right)}^{{2}}{y}{''}+{\left({x}+{1}\right)}{y}'+{y}={x}^{{2}}+{2}{\sin{{\left({\ln{{\left({x}+{1}\right)}}}\right)}}},{y}{\left({0}\right)}={\frac{{{1}}}{{{5}}}},{y}'{\left({0}\right)}={2}\)



Answered question


Differential equations

Answer & Explanation

Demetrius Kaufman

Demetrius Kaufman

Beginner2022-03-26Added 10 answers

Step 1
First of all , I found the equation solution of y+y(x+1)+y(x+1)2=0
I try to solve this ode using the variation of parameters theorem
Get this system equation:
1) c1cos(ln(x+1))+c2sin(ln(x+1))=0
2) c1sin(ln(x+1))+c2cos(ln(x+1))=x2+2sin(ln(x+1))
Multiply 1) by sin(ln(x+1)), 2) by cos(ln(x+1))x+1.
Step 2
By addtion i get:
I do not know how I get c2 by an integral ?


Beginner2022-03-27Added 13 answers

Step 1
The two equations you have using variation of parameters is incorrect. Given a second-order linear inhomogeneous DE y +p(x)y+q(x)y=f(x) with homogeneous solution yc(x)=c1y1(x)+c2y2(x), a particular solution yp is given by yp(x)=u1(x)y1(x)+u2(x)y2(x), where u1 and u2 satisfy

u 1 y 1 + u 2 y 2 = 0 u 1 y 1 + u 2 y 2 = f ( x ) .

Let y1=cos(ln(x+1))  and  y2=sin(ln(x+1)).

We have that
Step 2
Multiplying (1) by sin(ln(x+1)), (2) by (x+1)cos(ln(x+1)), and adding the two resulting equations, we get
u2 =x2+2sin(ln(x+1))x+1cos(ln(x+1)).
Step 3
Using the substitution z=ln(x+1), we see that
u2=[(ez1)2+2sinz]cosz, dz .
Computing this integral is a standard exercise now. Can you take it from here?

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?