# Use the exponential shift to find the general solution.

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Dutsky Kamdon Use the exponential shift to find the general solution.

1. (4D + 1)^4 y = 0.
2. (6D − 5)^3 y = 0.

fitzmerl duron 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 number 1, 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

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

Alternate solutions are encouraged.....

Sa ikalawang tanung....ang haba rin ng solution...pero parehas lang ng solution sa question no. 1.......Practice lang kayo......madali lang number 2.....parehas ang solution sa number 1....hehehehehe

Jhun Vert Wow... astig!

fitzmerl duron Thanks, Romel......pinagsikapan ko....sinuwerte nasagot....hehehehe

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