Homogeneous solution:
<span><span>y′′</span>+<span>y′</span>−6y=0</span>
yields the characteristic equation
<span><span>r2</span>+r−6=0 ⇒ r=−3,2</span>
So homogeneous part is <span><span>yc</span>=<span>C1</span><span>e<span>−3t</span></span>+<span>C2</span><span>e<span>2t</span></span></span>.
Non-homogeneous solution:
<span><span>y′′</span>+<span>y′</span>−6y=12<span>e<span>3t</span></span>+12<span>e<span>−2t</span></span></span>
Use the method of undetermined coefficients. Suppose <span><span>yp</span>=A<span>e<span>3t</span></span>+B<span>e<span>−2t</span></span></span> is a solution, so you have
<span><span><span><span> yp</span>=A<span>e<span>3t</span></span>+B<span>e<span>−2t</span></span></span><span><span>y′p</span>=3A<span>e<span>3t</span></span>−2B<span>e<span>−2t</span></span></span><span><span>y′p</span>=9A<span>e<span>3t</span></span>+4B<span>e<span>−2t</span></span></span></span></span>
Substitute into the original equation:
<span><span><span><span>(<span>9A<span>e<span>3t</span></span>+4B<span>e<span>−2t</span></span></span>)</span>+<span>(<span>3A<span>e<span>3t</span></span>−2B<span>e<span>−2t</span></span></span>)</span>−6<span>(<span>A<span>e<span>3t</span></span>+B<span>e<span>−2t</span></span></span>)</span></span><span>6A<span>e<span>3t</span></span>−4B<span>e<span>−2t</span></span></span></span><span><span>=12<span>e<span>3t</span></span>+12<span>e<span>−2t</span></span></span><span>=12<span>e<span>3t</span></span>+12<span>e<span>−2t</span></span></span></span></span>
This tells you that <span>A=2</span> and <span>B=−3</span>, and so your non-homogeneous part is <span><span>yp</span>=2<span>e<span>3t</span></span>−3<span>e<span>−2t</span></span></span>.
Your final solution would be the sum of the non/homogeneous parts, or
<span>y=<span>yc</span>+<span>y<span>p</span></span></span>
