Use the superposition approach to solve the non-homogeneous differential equation,y″+6y′+8y=4x−e−2x

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We have to use the superposition approach to solvethe non-homogeneous differential equations:-y&amp;nbsp;″+6y′+8y=4x−e−2xTake homogeneous eqn for Complemantary functiony&amp;nbsp;″+6y′+8y=0⇒(D2+6D+8)y=0The auxiliary equation ism2+6m+8=0⇒m2+4m+2m+8=0⇒m(m+4)+2(m+4)=0Step 2⇒(m+2)(m+4)=0⇒m=−2,−4∴yc(x)=c1e−2x+c2e−4xNow for particuler solution :-Let the particuler Solutionyp(x)=a1+a2x+a3e−2xx⇒yp′(x)=a2+a3e−2x−2a3e−2xx⇒y&amp;nbsp;″3(x)=−2a3e−2x+4a3e−2xx=−4a3e−2x+4a3e−2xx⇒−4a3e−2x+4a3e−2xx+6[a2+a3e−2x=2a3e−2xx]+8[a1+a2x+a3e−2xx]=4x−3+e−2x⇒e-2x(-4a3+4a3+6a3-12a3x+8a3x)+x(8a2)+(6a2+8a1)=-4x-3+e-2xEquating the coefficient of x,&amp;nbsp;e−2x&amp;nbsp;&amp;amp; constant⇒2a3=1⇒a3=138a2=4⇒a2=126a2+8a1=−3⇒6a1=−3−6a2⇒8a1=−3×13⇒6a1=−6⇒a1=−68⇒a1−3−4So,yp(x)=−34+12x+12e−2xxthen&amp;nbsp;y(x)=yc(x)+ypx⇒y(x)=c1e−2x+c2e−4x+12x+12e−2xx−−34⇒y(x)=(c1+x2)e−2x+c2e−4x+x2−−34

Use the superposition approach to solve the non-homogeneous

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