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Having trouble with this differential equation.

Find the general solution to the differential equation: $$ \frac {dy}{dx}= \frac {√y(x^2 + x − 4)} {(x^2 + 1)(x − 1)} $$

I don't know where to start.

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    $\begingroup$ The equation is separable... $\endgroup$
    – Ian
    May 21, 2016 at 12:38
  • 2
    $\begingroup$ is this a square root on the right hand side of the equation? $\endgroup$ May 21, 2016 at 12:38
  • $\begingroup$ how much of the numerator is under the square root? $\endgroup$ May 21, 2016 at 16:55

1 Answer 1

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$$\text{y}'\left(x\right)=\frac{\sqrt{\text{y}(x)}(x^2+x-4)}{(x^2+1)(x-1)}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x\tag1$$

Substitute $\text{u}=\text{y}\left(x\right)$ and $\text{d}\text{u}=\text{y}'\left(x\right)\space\text{d}x$ to find the integral $\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x$:

$$\int\frac{\text{y}'\left(x\right)}{\sqrt{\text{y}\left(x\right)}}\space\text{d}x=2\sqrt{\text{y}\left(x\right)}+\text{C}_1\tag2$$

For the RHS use partial fractions:

$$\frac{x^2+x-4}{(x^2+1)(x-1)}=\frac{2x}{x^2+1}+\frac{3}{x^2+1}-\frac{1}{x-1}\tag3$$

So, you get that:

$$\int\frac{x^2+x-4}{(x^2+1)(x-1)}\space\text{d}x=\ln\left|x^2+1\right|+3\arctan(x)-\ln\left|x-1\right|+\text{C}_2\tag4$$

So, we know that:

$$2\sqrt{\text{y}\left(x\right)}=\ln\left|x^2+1\right|+3\arctan(x)-\ln\left|x-1\right|+\text{C}\tag5$$

Solving for $\text{y}\left(x\right)$:

$$\text{y}\left(x\right)=\left\{\frac{\ln\left|\frac{x^2+1}{x-1}\right|+3\arctan(x)}{2}+\text{C}\right\}^2\tag6$$

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