# Getting the general solution of the differential equation $(4D+1)^4y=0$

I was looking for the solution of the differential equation $(4D+1)^4y=0$ using Exponential Shift Theorem.

My work:

The formula for getting a solution of a differential equation is $$P(D)(e^{rx}f(x)) = e^{rx}P(D+r)f(x)$$

With the form of differential equation given in the problem, the Exponential Shift Theorem formula is useless, so we need to modify the given differential equation so that we can use the Exponential Shift Theorem formula.

Now modifying the given differential equation:

$$(4D+1)^4y=0$$ $$\left(4\left(D+\frac{1}{4}\right)\right)^4y=0$$ $$4\left(D+\frac{1}{4}\right)^4y=0$$ $$\left(D+\frac{1}{4}\right)^4y=0$$

Then we let $u = e^{\frac{1}{4}x}y$, then getting the $y$: $y = e^{-\frac{1}{4}x}u$. So....

$$\left(D+\frac{1}{4}\right)^4y=0$$ $$\left(D+\frac{1}{4}\right)^4\left(e^{-\frac{1}{4}x}u\right)=0$$

Now....the form above is similar to the Exponential Shift Theorem form....Compare the two expressions below:

$$P(D)(e^{rx}f(x)) = e^{rx}P(D+r)f(x)$$ $$P(D)(e^{-\frac{1}{4}x}u)^4 = e^{-\frac{1}{4}x}P(D+\frac{1}{4})u$$

Then...we can now do this:

$$\left(D+\frac{1}{4}\right)^4y=0$$ $$\left(D+\frac{1}{4}\right)^4\left(e^{-\frac{1}{4}x}u\right)=0$$

Using the Exponential Shift Theorem, we can rewrite $\left(D+\frac{1}{4}\right)^4\left(e^{-\frac{1}{4}x}u\right)=0$ as:

$$\left(D+\frac{1}{4}\right)^4\left(e^{-\frac{1}{4}x}u\right)=0$$ $$e^{-\frac{1}{4}x} D^4u = 0$$ $$D^4u = 0$$

Now....get the integral of D^4 until only the $u$ remains.

$$D^4u = 0$$ $$\int D^4u = \int 0$$ $$D^3u + c_1 = c_2$$ $$D^3u = c_1$$

Next:

$$D^3u = c_1$$ $$\int D^3u = \int c_1$$ $$D^2u + c_2 = c_1x + c_3$$ $$D^2u = c_1x + c_2$$

Next:

$$D^2u = c_1x + c_2$$ $$\int D^2u = \int c_1x + \int c_2$$ $$Du + c_3 = c_1 \left(\frac{x^2}{2}\right) + c_2x + c_3$$ $$Du = c_1 x^2 + c_2x + c_3$$

Lastly:

$$Du = c_1 x^2 + c_2x + c_3$$ $$\int Du = \int c_1 x^2 + \int c_2x + \int c_3$$ $$u + c_4 = c_1 \left( \frac{x^3}{3}\right) + c_2 \left( \frac{x^2}{2}\right) + c_3x + c_4$$ $$u = c_1 x^3 + c_2x^2 + c_3x + c_4$$

We now got the $u$, so we almost got the solution to the given differential equation above:

$$y = e^{-\frac{1}{4}x} u$$ $$y = e^{-\frac{1}{4}x} (c_1 x^3 + c_2x^2 + c_3x + c_4)$$ $$y = c_1e^{-\frac{1}{4}x}x^3 + c_2e^{-\frac{1}{4}x}x^2 + c_3e^{-\frac{1}{4}x}x + c_4e^{-\frac{1}{4}x}$$

The solution to the differential equation above is

$$y = c_1e^{-\frac{1}{4}x}x^3 + c_2e^{-\frac{1}{4}x}x^2 + c_3e^{-\frac{1}{4}x}x + c_4e^{-\frac{1}{4}x}$$

I couldn't verify the answer, so I turn here...Is my answer correct?

• Yes. It's correct with many details. – Nosrati Nov 18 '17 at 3:41
• Char. equation is $(4\lambda+1)^4=0$ with solutions $\lambda_1=\lambda_2=\lambda_3=\lambda_4=-1/4$. Then your solution is correct. – Aleksas Domarkas Nov 13 '18 at 10:55