Final Answer
$\frac{-\left(y-1\right)^{3}}{3}=4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$
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Step-by-step Solution
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We could not solve this problem by using the method: Exact Differential Equation
1
Group the terms of the equation
$\frac{1}{\sqrt{x^2+4}}dy=-\left(\frac{1}{\left(y-1\right)^2}\right)dx$
2
Multiplying the fraction by $-1$
$\frac{1}{\sqrt{x^2+4}}dy=\frac{-1}{\left(y-1\right)^2}dx$
3
Divide fractions $\frac{1}{\frac{-1}{\left(y-1\right)^2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
$-\left(y-1\right)^2=\frac{1}{\frac{1}{\sqrt{x^2+4}}}$
4
Divide fractions $\frac{1}{\frac{1}{\sqrt{x^2+4}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$
$-\left(y-1\right)^2=\sqrt{x^2+4}$
5
Integrate both sides of the differential equation, the left side with respect to
$\int-\left(y^2-2y+1\right)dy=\int\sqrt{x^2+4}dx$
Intermediate steps
6
Solve the integral $\int-\left(y^2-2y+1\right)dy$ and replace the result in the differential equation
$\frac{-\left(y-1\right)^{3}}{3}=\int\sqrt{x^2+4}dx$
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Intermediate steps
7
Solve the integral $\int\sqrt{x^2+4}dx$ and replace the result in the differential equation
$\frac{-\left(y-1\right)^{3}}{3}=4\int\sec\left(\theta \right)^{3}d\theta$
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Intermediate steps
8
Solve the integral $4\int\sec\left(\theta \right)^{3}d\theta$ and replace the result in the differential equation
$\frac{-\left(y-1\right)^{3}}{3}=4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$
Explain this step further
Final Answer$\frac{-\left(y-1\right)^{3}}{3}=4\left(\frac{x\sqrt{x^2+4}}{8}+\frac{1}{2}\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)\right)+C_0$