Solve the initial value problem

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y″−2y′+y=δ(t−1),y(0)=2,y′(0)=3The general solution of the homogeneous equation is:yh(t)=c1et+c2tetTo find a particular solution of the inhomogeneous equation, we assume a function that satisfies the inhomogeneous term. In this case, we need to find a function that is zero for all t≠1 and infinity at t=1 such that the area under the function is equal to one. A function that satisfies these conditions is:yp(t)={0,t&amp;lt;1a,t&amp;gt;1where a is a constant to be determined.Using the definition of the Dirac delta function, we have:∫−∞∞δ(t−1)dt=1⇒∫−∞∞δ(t−1)f(t)dt=f(1)for any function f(t) that is continuous at t=1.Therefore, we have:∫−∞∞δ(t−1)yp(t)dt=yp(1)=aSince the inhomogeneous equation is linear, we can superpose the homogeneous and particular solutions to obtain the general solution:y(t)=yh(t)+yp(t)=c1et+c2tet+au(t−1)where u(t−1) is the unit step function, which is zero for all t&amp;lt;1 and one for all t&amp;gt;1.Using the initial conditions y(0)=2 and y′(0)=3, we can determine the values of c1, c2, and a as follows:c1+a=2c1+c2=3Differentiating y(t) with respect to t and evaluating at t=0 gives:y′(0)=c1+c2+aδ(0−1)=3Simplifying, we get:c1=1,c2=2,a=1Therefore, the solution to the initial value problem is:y(t)=et+2tet+u(t−1)

Solve the initial value problem

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