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91Ó°ÊÓ

Express the solution to the initial value problem y''-y=1t,y(1)=0,y'(0)=-2, using definite integrals. Using numerical integration (Appendix C) to approximate the integrals, find an approximation fory(2) to two decimal places.

Short Answer

Expert verified

The general solution isy(t)=e1-t-et-1+∫e-t2tdtet+∫-e-t2tdte-t and the approximation value for y(2)=-1.93.

Step by step solution

01

Find the particular solution

The differential equation is y''-y=1t.

The homogenous equation is r2-1=0.

Two independent solutions are r=±1.

Theny1=et,y2=te-t

yh(t)=c1et+c2e-t

The particular solution is yp=v1(t)et+v2(t)te-t.

02

Evaluate v1 and v2

Hereyp=v1(t)et+v2(t)te-t

And referring to (9) and solve the system then

v1'et+v2'e-t=0v1'et-v'2e-t=fav1'et-v'2e-t=1t

03

Find v1'and v1

v1'=-f(t)y2(t)ay1(t)y'2(t)-y'1(t)y2(t)=-(1t).e-tet.(-e-t)-et.e-t=-(1t).e-t-2=e-t2t

Now integrating this.

v1(t)=∫e-t2tdt

04

Determine v2' and v2

v2'=f(t)y1(t)ay1(t)y'2(t)-y'1(t)y2(t)=(1t).etet.(-e-t)-et.e-t=(1t).e-t-2=-e-t2t

Integrate this.

y2(t)=∫-e-t2tdt

Thus, a particular solution is:

yp=∫e-t2tdtet+∫-e-t2tdte-t

And the general solution is:

y(t)=yh(t)+yp(t)y(t)=c1et+c2e-t+∫e-t2tdtet+∫-e-t2tdte-t

05

Apply initial conditions

The given initial conditions are.

y(1)=c1e1+c2e-1+∫e-12dte1+∫-e-12dte-10=c1e+c2e-1c2=-c1e2

And

y'(t)=c1et-c2e-t+∫e-t2tdtet+12t+∫e-t2tdte-t-12ty'(1)=c1e-c2e-1+∫e-12dte+12+∫e-12dte-1-12-2=c1e-c2e-1

Solving for c1 and c2.

c1=-1e and c2=e

There for the solution is y(t)=e1-t-et-1+∫e-t2tdtet+∫-e-t2tdte-t.

And

y(2)=e1-2-e2-1+∫e-22(2)dte2+∫-e-22(2)dte-2=e-1-e=-1.93

This is the required result.

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