Final Answer
Step-by-step Solution
Specify the solving method
We could not solve this problem by using the method: Homogeneous Differential Equation
Group the terms of the equation
Multiplying the fraction by $-1$
Multiply both sides of the equation by $\sqrt{x^2+4}$
Divide both sides of the equation by $d
Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$
The differential equation $\left(y-1\right)^2dy1\sqrt{x^2+4}dx=0$ is exact, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and they satisfy the test for exactness: $\displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form $f(x,y)=C$
Using the test for exactness, we check that the differential equation is exact
Integrate $M(x,y)$ with respect to $x$ to get
Now take the partial derivative of $\frac{1}{2}\sqrt{x^2+4}x+2\ln\left(\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right)$ with respect to $y$ to get
Set $\left(y-1\right)^2$ and $0+g'(y)$ equal to each other and isolate $g'(y)$
Find $g(y)$ integrating both sides
We have found our $f(x,y)$ and it equals
Then, the solution to the differential equation is
Group the terms of the equation