Question #a16d8

Dec 19, 2016

The solution is $y = x - 8 \ln | x | + 1$

Explanation:

$\mathrm{dy} = \frac{x - 8}{x} \mathrm{dx}$

$\mathrm{dy} = \left(x - 8\right) {x}^{-} 1 \mathrm{dx}$

$\int \left(\mathrm{dy}\right) = \int \left(x - 8\right) {x}^{-} 1 \mathrm{dx}$

Integrate by parts on the right-hand side. Let $\mathrm{dv} = \frac{1}{x}$ and $u = x - 8$. Then $v = \ln | x |$ and $\mathrm{du} = 1 \mathrm{dx}$.

The integration by parts formula states that $\int \left(u \mathrm{dv}\right) = u v - \int \left(v \mathrm{du}\right)$.

Therefore:

$\int \left(x - 8\right) {x}^{-} 1 = \left(x - 8\right) \ln | x | - \int \left(\ln | x |\right)$

We will now need to reintegrate using integration by parts.

We let $u = \ln | x |$ and $\mathrm{dv} = 1$. $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = x$.

$\int \left(\ln | x |\right) = x \ln | x | - \int \left(\frac{1}{x} \times x\right)$

$\text{ } = x \ln | x | - x$

$\text{ } = x \left(\ln | x | - 1\right) + C$

So, the complete integration, after integrating both sides, will be:

$y = \left(x - 8\right) \ln | x | - \left(x \left(\ln | x | - 1\right)\right) + C$

$y = \left(x - 8\right) \ln | x | - x \ln | x | + x + C$

$y = x \ln | x | - 8 \ln | x | - x \ln | x | + x + C$

$y = x - 8 \ln | x | + C$

We want the equation that passes through $\left(1 , 2\right)$. Thus:

$2 = \left(1 - 8\right) \ln | 1 | - 1 \left(\ln | 1 |\right) + 1 + C$

$2 = - 7 \left(0\right) - 1 \left(0\right) + 1 + C$

$2 - 1 = C$

$C = 1$

The solution to the differential equation is therefore $y = x - 8 \ln | x | + 1$

Hopefully this helps!