Use the laplace transform to solve y"-2y'+2y=cos(t); y(0)=1, y'(0)=0.

y'' - 2∙y' + 2∙y = cos(t)
=>
ℒ{ y'' - 2∙y' + 2∙y } = ℒ{ cos(t) }
<=>
ℒ{ y'' } - 2∙ℒ{ y' } + 2∙ℒ{ y } = ℒ{ cos(t) }
<=>
s²∙ℒ{ y } - s∙y(0) - y'(0) - 2∙(s∙ℒ{ y } - y(0)) + 2∙ℒ{ y } = s/(s² + 1)
<=>
s²∙ℒ{ y } - s - 2∙(s∙ℒ{ y } - 1) + 2∙ℒ{ y } = s/(s² + 1)
<=>
(s² - 2∙s + 2)∙ℒ{ y } = (s/(s² + 1)) + (s - 2)
<=>
ℒ{ y } = [s/(s² + 1)∙(s² - 2∙s + 2)] + [(s - 2)/(s² - 2∙s + 2)]

Decomposition to partial fractions for the first term on right hand side:
s/(s² + 1)∙(s² - 2∙s + 2) = (A∙s + B)/(s² + 1) + (C∙s + D)/(s² - 2∙s + 2)
<=>
s = (A∙s + B)∙(s² - 2∙s + 2) + (C∙s + D)∙(s² + 1)
<=>
s = s³∙(A + C) + s²∙(-2∙A + B + D) + s∙(2∙A - 2∙B + C) + 1∙(2∙B + D)
comparison of coefficients of same powers of s leads to
A + C = 0
-2∙A + B + D = 0
2∙A - 2∙B + C = 1
2∙B + D = 0
Solution to this equation system is:
A = 1/5
B = -2/5
C = -1/5
D = 4/5

Hence,
ℒ{ y } = (1/5)∙[(s - 2)/(s² + 1)] - (1/5)∙[(s - 4)/(s² - 2∙s + 2)] + [(s - 2)/(s² - 2∙s + 2)]
= (1/5)∙[(s - 2)/(s² + 1)] + (1/5)∙[(4∙s - 6)/(s² - 2∙s + 2)]
=>
y = ℒ⁻¹{ (1/5) ∙ [(s - 2)/(s² + 1)] + (1/5)∙[(4∙s - 6)/(s² - 2∙s + 2)] }
= (1/5)∙( ℒ⁻¹{ (s - 2)/(s² + 1) } + ℒ⁻¹{ (4∙s - 6)/(s² - 2∙s + 2) } )

The first term can be easily expanded into the transforms of a sine and cosine function:
ℒ⁻¹{ (s - 2)/(s² + 1) }
= ℒ⁻¹{ s/(s² + 1) } - 2∙ℒ⁻¹{ 1/(s² + 1) }
= cos(t) - 2∙sin(t)

The second term ra be reduced to the transform of sine and cosine by applying the shifting theorem
ℒ⁻¹{ F(s - a) } = e^(a∙t)∙ℒ⁻¹{ F(s) }
=>
ℒ⁻¹{ (4∙s - 6)/(s² - 2∙s + 2) }
= ℒ⁻¹{ (4∙s - 4 - 2)/(s² - 2∙s + 1 + 1) }
= ℒ⁻¹{ (4∙(s - 1) + 2)/( (s - 1)² + 1) }
= e^(t) ∙ ℒ⁻¹{ (4∙s + 2)/( s² + 1) }
= e^(t) ∙ ( 4∙ℒ⁻¹{ s/(s² + 1) } - 2∙ℒ⁻¹{ 1/(s² + 1) }
= e^(t) ∙( 4∙cos(t) - 2∙sin(t) )

=>

y(t) = (1/5)∙( cos(t) - 2∙sin(t) + e^(t)∙(4∙cos(t) - 2∙sin(t)) )

Definitely