# Thread: General Solution of Differenatial Equation

1. ## General Solution of Differenatial Equation

I need help with finding the general solution of this differenatial equation:

1/x^6 dy/dx = 19 + 9y

Thanks

2. ## Re: General Solution of Differenatial Equation

Hello, MMCS!

Separate the variables . . . and integrate.

3. ## Re: General Solution of Differenatial Equation

Multiply by $x^6$

$\frac{dy}{dx}=x^6(19+9y)$

Divide by $19+9y$

$\frac{dy}{dx(19+9y)}=x^6$

Integrate

$\int\frac{dy}{19+9y}=\int x^6 dx$

$\int \frac{dy}{19+9y}=\int\frac{dy}{9(\frac{19}{9}+y)}= \frac{1}{9}\ln(\frac{19}{9}+y)+C$

$\int x^6dx=\frac{x^7}{7}+\bar{C}$

$\frac{1}{9}\ln(\frac{19}{9}+y)+C=\frac{x^7}{7} + \bar{C} \rightarrow \ln(\frac{19}{9}+y)=\frac{9x^7}{7}+D$

Taking exponentials

$\frac{19}{9}+y=e^{\frac{9x^7}{7}+D}=Ke^{\frac{9x^7 }{7}}$

So

$\boxed{y=Ke^{\frac{9x^7}{7}}-\frac{19}{9}, K \in \mathbb{R}}$

$\frac{dy}{dx}=Ke^{\frac{9x^7}{7}}9x^6$

Noting that

$Ke^{\frac{9x^7}{7}}=y+\frac{19}{9}$

$\frac{dy}{dx}=\left(y+\frac{19}{9}\right)9x^6=(9y+ 19)x^6$

As requested