Question

Muu paldmeter.
Solve (D2 - 2D + 1)y = cos3x.​

Answer 1

Therefore the Answer is $GE=(A+Bx)e^{x} +\frac{3x^{2}e^{x} }{4} +\frac{3e^{-x} }{8}$

Step-by-step explanation:

Given;

Differential equation is $(D^{2} -2D+1)y=cos3x$

The auxiliary equation is

$m^{2} -2m+1=0$

Also write;

$m^{2} -2\times m \times 1+1^{2} =0$

$(m-1)^{2} =0$   ( ∵ $a^{2} -2ab+b^{2} =(a-b)^{2}$ )

$m=1,1$  

The root are real and equal.

So the complementary function is

$CF=(A+Bx)e^{x}$

Particular  Integral,

$f(x)=e^{ax+b}$

Now we have to find the particular integral,

$PI=\frac{cos3x}{D^{2}-2D+1 }$$=\frac{3(e^{x} +e^{-x} )}{2(D^{2}-2D+1) }$$=PI_{1} +PI_{2}$

Now,

$PI_{1} =\frac{3e^{x} }{2(D^{2}-2D+1) }$

Plug $D=1$ in above equation,

$PI_{1} =\frac{3e^{x} }{2(1^{2}-2\times1+1) } =$ undefined

$PI_{1} =\frac{3x^{2} e^{x} }{4}$

Similarly,

$PI_{2}= \frac{3e^{-x} }{2((-1)^{2}-(2\times-1)+1) }$$=\frac{3e^{-x} }{8}$    ( By substituting $D=-1$ )

Now the general solution is,

$GE=CF+PI_{1} +PI_{2}$

∴  $GE=(A+Bx)e^{x} +\frac{3x^{2}e^{x} }{4} +\frac{3e^{-x} }{8}$

So the Answer is $GE=(A+Bx)e^{x} +\frac{3x^{2}e^{x} }{4} +\frac{3e^{-x} }{8}$

Answer 2

Answer:

Step-by-step explanation: