How do you solve this differential equation using variation of parameters?


$\color{green}{I~tried}$ . . .

$using~the~\color{blue}{Laplace~transform}~method$ . . .

$$L[\int_{0}^{\infty }\frac{sinxt}{1+t^{2}}dt]$$

$$=\int_{0}^{\infty }e^{-sx}(\int_{0}^{\infty }\frac{sinxt}{1+t^{2}}dt)$$

$$=\int_{0}^{\infty }\frac{1}{1+t^{2}}(\int_{0}^{\infty }e^{-px}sinxtdx)dt\\\\\\=\int_{0}^{\infty }\frac{1}{1+t^{2}}\frac{t}{s^{2}+t^{2}}dt$$

$$=\int_{0 }^{\infty }\frac{1}{s^{2}-1}(\frac{t}{1+t^{2}}-\frac{t}{s^{2}+t^{2}})$$


Is my solution correct?

Should I use the inverse Laplace?

How can I get a complete and correct answer?

Thanks for any hint.

  • 1
    $\begingroup$ You're welcome! It's always a pleasure to encounter posts where thought and effort are clearly shown! +1 $\endgroup$ – Namaste Aug 27 '13 at 12:16
  • $\begingroup$ @amWhy thank you again :) $\endgroup$ – Software Aug 27 '13 at 12:19

I will map out the steps for you and have you fill in the details.

We are asked to solve this using Variation of Parameters (VoP), given:

$$\tag 1 y''-\dfrac{2x}{x^2+1}y'+\dfrac{2}{x^2+1}y=6(x^2+1)$$

Step 1

Find the homogenous solution to $(1)$, so we have:

$$\tag 2 y''-\dfrac{2x}{x^2+1}y'+\dfrac{2}{x^2+1}y=0$$

This yields:

$$y_h = c_1(x^2-1) + c_2 x$$

Step 2

We are now going to make use of VoP, so we set: $y_1 = x^2-1$ and $y_2 = x$ from $y_h$ and $f = 6(x^2+1)$ from $(1)$.

We calculate the Wronskian of $y_1$ and $y_2$, yielding $W(x^2-1, x) = -x^2-1$.

Using VoP, we have:

$$u_1 = \int \dfrac{-y_2 f}{W(x^2-1, x)} dx = \int \dfrac{-x 6(x^2+1)}{-x^2-1} dx = 3x^2$$

$$u_2 = \int \dfrac{y_1 f}{W(x^2-1, x)} dx = \int \dfrac{(x^2-1)6(x^2+1)}{-x^2-1} dx = 6x-2x^3$$

Now, $y_p$ is given by:

$$y_p = y_1 u_1 + y_2 u_2 = (x^2-1)(3x^2) + (-2x^3+6x)(x) = x^4 + 3x^2$$

Step 3

Our final solution is given by:

$$y(x) = y_h(x) + y_p(x) = c_1(x^2-1) + c_2 x + x^4 + 3x^2$$

  • $\begingroup$ Your advice is excellent. Thank you @Amzoti $\Large\color{green}{✓}^+$ $\endgroup$ – Software Aug 24 '13 at 7:15
  • $\begingroup$ Your guide was very helpful. I understand completely.I realized that I was wrong. $\endgroup$ – Software Aug 24 '13 at 7:22
  • $\begingroup$ can I use Laplace transform?is It impossible?Do I need to do?? $\endgroup$ – Software Aug 24 '13 at 7:34
  • $\begingroup$ Yes, It is true. Thanks for your tips and answers. :) $\endgroup$ – Software Aug 24 '13 at 16:48
  • $\begingroup$ Nice, step-by-step!! +1 $\endgroup$ – Namaste Aug 25 '13 at 0:31

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