Question

Solve : (d^3y÷dx^3)‐7(dy÷dx)+6y=e^x

Answer 1

Answer:

Will depend on values of y(0), y'(0), y''(0).

Step-by-step explanation:

You can use Laplace transform to solve this:

$y'''-7y'+6y=e^x$

Taking Laplace transform on both sides:

$\mathcal{L}(y'''-7y'+6y)=\mathcal{L}(e^x)\\s^3\mathcal{L}(y)-s^2y(0)-sy'(0)-y''(0)-7s\mathcal{L}(y)+7y(0)+6\mathcal{L}(y)=\frac{1}{s-1}$

Now just try to make rearrangements to get the terms in better way for inverse laplace.

$(s^3-7s+6)\mathcal{L}(y)=s^2y(0)+sy'(0)+y''(0)-7y(0)+\frac{1}{s-1}\\\mathcal{L}(y)=\frac{as^2+bs+c-7a}{s^3-7s+6}+\frac{1}{(s-1)(s^3-7s+6)}$

Now the partial fractions of the two terms can be done and use the following to convert by laplace inverse on both sides.

$\mathcal{L}(e^{at})=\frac{1}{s-a}\\\mathcal{L}(\sin (at))=\frac{a}{s^2+a^2}\\\mathcal{L}(\cos(at))=\frac{s}{a^2+s^2}$

The partial fractions can be done using the data of y(0), y'(0), y''(0) or by considering them as arbitrary constants.

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