**Problem Statement:**Use the Laplace transform to solve the given initial value problem.\[ y'' - 2y' + 2y = \cos t; \quad y(0) = 1, \quad y'(0) = 0 \]**Solution:**\[ y(t) = \boxed{} \] **Instructions:**To solve this initial value problem using the Laplace transform:1. **Apply the Laplace Transform** to both sides of the differential equation.2. Utilize the initial conditions to solve for \( Y(s) \) in the Laplace domain.3. **Find the Inverse Laplace Transform** of \( Y(s) \) to obtain the solution \( y(t) \) in the time domain.This method allows for solving linear differential equations by transforming them into algebraic equations, simplifying the process.

The differential equation is given as : y''-2y'+2y=cos ty(0)=1y'(0)=0 The objective is to find the Laplace transform of the given IVP. Apply the Laplace transform on both sides of the given equation. L{y''}-2L{y'}+2L{y}=L{cos t} ............(1) Calculate the value for L{y'} , L{y'}=sL{y}-y(0)=sL{y}-1 And the value for L{y''} , L{y''}=s2L{y}-sy(0)-y'(0)=s2L{y}-s(1)-0=s2L{y}-s And the value of L{cos t} L{cos t}=ss2+1Use these expressions, in equation (1). L{y''}-2L{y'}+2L{y}=L{cos t}s2L{y}-s-2sL{y}-1+2L{y}=ss2+1s2L{y}-s-2sL{y}+2+2L{y}=ss2+1s2-2s+2L{y}-s+2=ss2+1s2-2s+2L{y}=ss2+1+s-2s2-2s+2L{y}=s3-2s2+2s-2s2+1L{y}=s3-2s2+2s-2s2+1s2-2s+2By the method of partial fractions, Y(s)=s3-2s2+2s-2s2-2s+2s2+1=as+bs2-2s+2+cs+ds2+1.........(2) Compare the numerators on both sides. s3-2s2+2s-2=as+bs2+1+cs+ds2-2s+2s3-2s2+2s-2=a+cs3+b-2c+ds2+a+2c-2ds+b+2d On equating the coefficients of like terms on both sides, we will obtain that a+c=1-2c+d=-2a+2c-2d=2b+2d=-2Now, solve the above equations as follows: Write the matrix form of the above equations. 101001-21102-20102abcd=1-22-2 The augmented matrix is, 101001-21102-20102:1-22-2 Apply the row operation, R3→R3-R1 on the above matrix. 101001-21001-20102:1-21-2 Apply the row operation, R4→R4-R2 on the above matrix. 101001-21001-20021:1-210 Apply the row operation, R4→R4-2R3 on the above matrix. 101001-21001-20005:1-21-2 Apply the row operation, R4→15R4 on the above matrix. 101001-21001-20001:1-21-25Now, convert the above matrix into system of equations as follows: a+c=1b-2c+d=-2c-2d=1d=-25 Substitute d=-25in the equation c-2d=1. c-2-25=1c+45=1c=1-45c=15 Substitute c=15 in the equation, a+c=1. a+15=1a=1-15a=45 Substitute c=15 ,d=-25 in the equation b-2c+d=-2. b-215+-25=-2b-25-25=-2b-45=-2b=-65 So, the obtained values are, a=45b=-65c=15d=-25Now, the partial fraction decomposition is, Y(s)=s3-2s2+2s-2s2-2s+2s2+1=as+bs2-2s+2+cs+ds2+1=45s+-65s2-2s+2+15s+-25s2+1=4s-65s2-2s+2+s-25s2+1 Substituting these values in equation (2), we obtain that Y(s)=4s-65s2-2s+2+s5s2+1-25s2+1=4s-1-25s-12+1+s5s2+1-25s2+1=45s-1s-12+1-251s-12+1+15ss2+1-251s2+1.......(3)Note that, the formula are : L-1s-1s-12+1=etcost L-11s-12+1=etsin t L-1ss2+1=cost L-11s2+1=sin t Then use linearity of Laplace transform, then equation (3) becomes, L-1Y(s)=45L-1s-1s-12+1-25L-11s-12+1+15L-1ss2+1-25L-11s2+1=45etcos t-25etsin t+15cos t-25sin tCONCLUSION : Therefore, the required solution is : y=45etcos t-25etsin t+15cos t-25sin t.