y"+3y'+2y=2 if 0&lt;t&lt;5 y(0)=0 and y'(0)=0

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y&quot;+3y&#039;+2y=2 if 0&amp;lt;t&amp;lt;5 y(0)=0 and y&#039;(0)=0

alenahelenash · Accepted Answer

The given differential equation is:y+3y′+2y=2To solve this, we can first find the homogeneous solution by solving the related homogeneous equation:yh+3yh′+2yh=0The characteristic equation is:r2+3r+2=0Factoring, we get:(r+1)(r+2)=0Therefore, the roots are r=−1 and r=−2. The homogeneous solution is:yh(t)=c1e−t+c2e−2twhere c1 and c2 are constants determined by the initial conditions.Next, we need to find a particular solution to the non-homogeneous equation. A guess for a particular solution is:yp(t)=Awhere A is a constant to be determined.Substituting this into the differential equation, we get:A+3(0)+2A=2Solving for A, we get:A=1Therefore, the particular solution is:yp(t)=1The general solution to the non-homogeneous equation is the sum of the homogeneous and particular solutions:y(t)=yh(t)+yp(t)=c1e−t+c2e−2t+1Using the initial condition y(0)=0, we get:c1+c2+1=0Using the initial condition y′(0)=0, we get:−c1−2c2=0Solving this system of equations, we get:c1=23 and c2=−53Therefore, the solution to the differential equation is:y(t)=23e−t−53e−2t+1 for 0&amp;lt;t&amp;lt;5.y(0)=23−53+1=0 and y′(0)=−23+103=0, so the initial conditions are satisfied.

y"+3y'+2y=2 if 0<t<5 y(0)=0 and y'(0)=0

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