Best Answer

In order to solve this inhomogeneous differential equation you need to start by solving the homogeneous case first (aka when the right hand side is just 0).

The characteristic equation for this differential equation is r²+1=0 or r²=-1 which means that r must be equal to ±i. Therefore, the general solution to this homogeneous problem Is y=c1*sin(x)+c2*cos(x) where c1 and c2 are constants determined by the initial conditions.

In order to solve the inhomogeneous problem we need to first find the Wronskian of our two solutions.

_________|y1(x) y2(x) | __| sin(x) cos(x) |

W(y1, y2)= |y1'(x) y2'(x) | = | cos(x) -sin(x) | = -sin(x)²-cos(x)²= -1

Next, we calculate the particular solution

Y(x)=-sin(x)* Integral(-1*cos(x)*cot(x)) + cos(x)*Integral(-1*sin(x)*cot(x))

=sin(x)*Integral(cos²(x)/sin(x)) - cos*Integral(cos(x))

=sin(x)*(ln(tan(x/2)) + cos(x)) -cos(x)*sin(x)=sin(x)*ln(tan(x/2))

Finally, to answer the entire equation, we add the particular solution to the homogeneous solution to get

y(x)=sin(x)*ln(tan(x/2)) + c1*sin(x)+c2*cos(x)

Q: Y'' plus y equals cotx diff eq?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

2x - 3y = -1 Eq.1y = -4x + 24 Eq.2Put the value of Eq.2 on Eq.22x - 3(-4x + 24 ) = -12x + 12x -72 = -114x -72 = -114x = 72-114x = 71x= 71/14 .............. X = 5.07

Eq 1: + y= 5 Eq 2: x = y + 1 substitute y + 1 for x in equqtion 1 y+1 + y = 5 2y + 1 = 5 2 y = 4 y = 2 substitute 2 for y in eq 2 x = 2 + 1 x = 3

3x + 6y = 48 .... Eq.1-5x + 6y = 32 .... Eq.2 .... in another way 6y = 5x + 32Put the value of Eq.2 in Eq. 13x + 5x + 32 = 488x = 48 - 328x = 16x = 16/8 ...... X = 2To find the value of Y put the value of X in Eq.26y = 5x + 326y = 5(2) + 326y = 10 + 326y = 42y = 42/6 ...... Y + 7

one solution write down the 2 equations eq 1: 2x + y = 5 eq 2: 2y + x = 4 get a value for x from eq 2 x = 4 - 2y substitute 4 - 2y for x in eq 1 2(4 - 2y) + y = 5 8 - 4y +y = 5 -3y = -3 y=1 substitute 1 for y in eq 2 2(1) + x = 4 2 + x = 4 x = 4-2 x = 2

5x + y = 5 .... Eq.1 we can write it in another way ... y = 5 - 5x3x + 2y = 3 .... Eq.2Put the value of Y in Eq.23x + 2(5 - 5x) = 33x + 10 - 10x = 33x - 10x = 3 - 10- 7x = -7x = -7/-7 ............. X = 1to find the value of y, put the value of x in Eq.1y= 5 - 5xy= 5 - 5(1)y= 5 - 5 ............. Y = Zero

Related questions

2x - 3y = -1 Eq.1y = -4x + 24 Eq.2Put the value of Eq.2 on Eq.22x - 3(-4x + 24 ) = -12x + 12x -72 = -114x -72 = -114x = 72-114x = 71x= 71/14 .............. X = 5.07

4x + 3y = -5 (Eq 1); 3x + 5y = -12 (Eq 2) Eliminate x Eq 1 times 3: 12x + 9y = -15 Eq 2 times 4: 12x +20y = -48 Subtract Eq 1 from Eq 2 11y = -48-(-15) ie 11y = -33 y = -3 Substitute in Eq 1: 4x - 9 = -5, ie 4x = 4 so x = 1 Check in Eq 2: (3 x 1) + (5 x -3) = 3 - 15 = -12. QED

Eq 1: + y= 5 Eq 2: x = y + 1 substitute y + 1 for x in equqtion 1 y+1 + y = 5 2y + 1 = 5 2 y = 4 y = 2 substitute 2 for y in eq 2 x = 2 + 1 x = 3

3x + 6y = 48 .... Eq.1-5x + 6y = 32 .... Eq.2 .... in another way 6y = 5x + 32Put the value of Eq.2 in Eq. 13x + 5x + 32 = 488x = 48 - 328x = 16x = 16/8 ...... X = 2To find the value of Y put the value of X in Eq.26y = 5x + 326y = 5(2) + 326y = 10 + 326y = 42y = 42/6 ...... Y + 7

one solution write down the 2 equations eq 1: 2x + y = 5 eq 2: 2y + x = 4 get a value for x from eq 2 x = 4 - 2y substitute 4 - 2y for x in eq 1 2(4 - 2y) + y = 5 8 - 4y +y = 5 -3y = -3 y=1 substitute 1 for y in eq 2 2(1) + x = 4 2 + x = 4 x = 4-2 x = 2

4x + 3y = -5 (Eq 1) 3x + 5y = -12 (Eq 2) Multiply the first equation by 3 and the second by 4: 12x + 9y = -15 (Eq 3) 12x + 20y = -48 (Eq 4) subtract Eq 3 from Eq 4 gives 11y = -33 so y = -3 and x = (15 + 12)/3 =1 Check: (4 x 1) + (3 x -3) = 4 - 9 = -5 and (3 x 1) + (5 x -3) = 3 - 15 = -12 QED Subtract

5x + y = 5 .... Eq.1 we can write it in another way ... y = 5 - 5x3x + 2y = 3 .... Eq.2Put the value of Y in Eq.23x + 2(5 - 5x) = 33x + 10 - 10x = 33x - 10x = 3 - 10- 7x = -7x = -7/-7 ............. X = 1to find the value of y, put the value of x in Eq.1y= 5 - 5xy= 5 - 5(1)y= 5 - 5 ............. Y = Zero

16

see if 4+4=3 then value of 3=4+4 (eq 1) now we need to find out 9+9=? we can write 9+9 as 3+3+3+3+3+3 (eq 2) now by substituting the value of 3 from eq 1 in eq 2 we will get (4+4)+(4+4)+(4+4)+(4+4)+(4+4)+(4+4)=48 so 9+9=48

3x + 2y = 4 (Eq 1); 4x + 3y = 7 (Eq 2) Eliminate x Eq 1 x 4 = 12x + 8y = 16 (Eq 3) Eq 2 x 3 = 12x + 9y = 21 (Eq 4) Subtract Eq 3 from Eq 4 y = 5 so x = (4 - 10)/3 = -6/3 ie -2 Check Eq 2: (4 x -2) + (3 x 5) = -8 + 15 = 7 QED

Eq 1: 3x + y = 10 Eq 2: 3x - 2y = -5 Subtract 1 from 2: -3y = -15 Divide by -3: y = 5 Substitute y = 5 in Eq 1: 3x + 5 = 10 so x = 5/3 Confirm in Eq 2: 3 x 5/3 - 10 = 15/3 - 10 = 5 - 10 = -5. QED

x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...