((d^2y)/(dx^2)) + (k(dy/dx)) + 9y = 36sin(3x) under the conditions y=0, dy/dx =3 when x=0 (i)if k=6 (ii)if k=0

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

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Given differential equation is `(d^2y)/(dx^2)+kdy/dx+9y=36sin3x`

and the initial conditions are

                            `y(0)=0`  and `dy/dx=3` at `x=0` .

(i) when k=6 the above differential equation becomes

                    `(d^2y)/(dx^2)+6dy/dx+9y=36sin3x`

Now the auxiliary equation is   `m^2+6m+9=0`

or,        m=-3,-3.

So the complementary function is  C.F.=`(A+Bx)e^-(3x)` .

P.I.=`(36sin3x)/(D^2+6D+9)`       Replacing D^2 by  -(3)^2 we get,

  =`(36sin3x)/(6D)` =-2cos3x.

So the solution is  y=(A+Bx)e^(-3x)-2cos3x.

Applying  y(0)=0

we get  A=2.

Applying  at x=0  dy/dx=3

we get B=9.

So the solution is y=(2+9x)e^(-3x)-2cos3x.

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oldnick | (Level 1) Valedictorian

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`2) k=0`      `lambda^2+9=0`     `lambda =+-3i`

In this case general integral is of the form:

 `c_1cos3x+c_2sin3x`

Developing particular integral  as above:

`y(x)=c_1cos3x+c_2sin3x-2sin3x`

With initial  conditions: `y(0)=0;y'(0)=3`  we get:

`c_1=0; c_2=5`  

So particular integral is:

`y(x)= ` `5sin3x`

 

 

 

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oldnick | (Level 1) Valedictorian

Posted on

 we can re-write equation as:

`y''+ky'+9y=36sin3x`

It's a second order non omogeneus linear equation. with inititial contidition:

`y(0)=0; y'(0)=0`

`1) k=6`

then equation becames:

`y''+6y'+9y= 36sin3x`   (1)

calvcolate  roots chatarterristics equation as it was omogeneuos

`lambda^2+6 lambda+9=0` `rArr`  `(lambda+3)^2=0`   that gives a double solution `lambda= -3` 

the solution of omoogeneous equation is:

`y(x)=c_1e^(-3x)+c_2xe^(-3x)`   (general integral)

since isn't omogeneous we have to get as new function:

 

`y(x)= Asin3x +Bcos3x+C`

`y'(x)= 3Acos3x-3Bsin3x```

`y''(x)=-9Asin3x-9Bcos3x `

 

So applied in (1) we get:

`-9Asin3x-9Bcos3x+18Acos3x-18Bsin3x+9Asin3x+9Bcos3x+9C=36sin3x `

```-18Bsin3x+18Acos3x+9C=36sin3x` 

so: `Acos3x-Bsin3x+C/2=2sin3x`

that shows:

`A=0;B=-2;C=0`

`y(x)=c_1e^(-3x)+c_2xe^(-3x)-2sin3x`  (particular integral)

since `y(0)= 0;y'(0)=0`   `rArr c_1=0;c_2=0`

Then:   `y=-2sin3x`

 

 

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