The given equation and initial values are $$y^{'''}+12y^{''}+36y^{'}=0$$ $$y(0)=0$$ $$y'(0)=1$$ $$y''(0)=-7$$ Using the auxiliary equation and factor we get $$m_1=0, \space m_{2,3}=-6$$ Then the equation becomes $$y= c_1+c_2e^{-6x}+c_3xe^{-6x}$$ Now taking the derivatives and substituting in the IVTs $$0=c_1+c_2$$ $$1=-6c_2+c_3$$ $$-7=36c_2-12c_3$$

Now here is where I am stuck, what are the steps in finding the values for $c_1, c_2,c_3$. I tried adding them all together but that was fruitless. I then tried to solve each equation independently in terms of each other but could find a proper fit.

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    $\begingroup$ Solve the last two equations alone as a system of two equations...Find $c_2,c_3$ then deduce $c_1=-c_2$ $\endgroup$ Commented Mar 30, 2020 at 18:47

1 Answer 1


From $$1=-6c_2+c_3\\ -7=36c_2-12c_3$$ we have $$6=-36c_2+6c_3\\-7=36c_2-12c_3$$ adding: $$-1=-6c_3\implies c_3=\frac{1}{6}.$$ Now from $1=-6c_2+c_3$, we have $$1=-6c_2+\frac{1}{6}\implies c_2=-\frac{5}{36} $$ and from $0=c_1+c_2$, we have $$c_1=-c_2=\frac{5}{36}.$$


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