MHB Substitution to solve fourth order diff equation

[Sudharaka]

pete078421's question from Math Help Forum,

Any help would wonderfully appreciated!

Find the General Solution of, \(\displaystyle\frac{d^4y}{dt^4}+2\frac{d^2y}{dt^2}=4t^2\) by using the substitution \(\displaystyle z(t)=\frac{d^{2}y}{dt^2}\)

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Hi pete078421,

Using the given substitution you get,

\[\frac{d^2z}{dt^2}+2z=4t^2\]

First we shall find the complementary function for this differential equation. The auxiliary equation is,

\[m^2+2=0\]

\[\Rightarrow m=\pm i\sqrt{2}\]

Therefore the complementary function is,

\[z_{c}(t)=Ae^{i\sqrt{2}t}+Be^{-i\sqrt{2}t}\mbox{ where }A\mbox{ and }B\mbox{ are arbitrary constants.}\]

Now we shall seek for the particular integral.

\[\frac{d^2z}{dt^2}+2z=4t^2\]

Using differential operator notations we can write this as,

\[(D^{2}+2)z=4t^2\]

\[\Rightarrow z=\left(\frac{1}{D^{2}+2}\right)4t^2\]

\[\Rightarrow z=\frac{1}{2}\left(1-\frac{D^2}{2}+\left(\frac{D^2}{2}\right)^2-\left(\frac{D^2}{2}\right)^3+\cdots\right)4t^2\]

Therefore the particular integral is,

\[z_{p}(t)=2t^{2}-2\]

The general solution of the differential equation is,

\[z(t)=z_{p}(t)+z_{c}(t)\]

\[\therefore z(t)=2t^{2}-2+Ae^{i\sqrt{2}t}+Be^{-i\sqrt{2}t}\mbox{ where }A\mbox{ and }B\mbox{ are arbitrary constants.}\]

Since, \(\displaystyle z(t)=\frac{d^{2}y}{dt^2}\) we get,

\[\frac{d^{2}y}{dt^2}=2t^{2}-2+Ae^{i\sqrt{2}t}+Be^{-i\sqrt{2}t}\]

Integrating twice we get,

\[y(t)=\frac{t^4}{6}-t^2+C_1e^{i\sqrt{2}t}+C_2e^{-i\sqrt{2}t}+C_3\mbox{ where }C_1,\,C_2\mbox{ and }C_3\mbox{ are arbitrary constants.}\]

 

[Ackbach]

I wonder if pete078421 might prefer answers in real functions. That is, use $z_{c}=A\cos\left(\sqrt{2}\,t\right)+B\sin\left( \sqrt{2}\,t\right).$