Answer:
![y = √(x^2+4)](https://img.qammunity.org/2021/formulas/mathematics/college/2om474v58zbi1ry4106yhf7mex3kaxe20e.png)
Explanation:
The question is incomplete. Here is the complete question.
Find the solution of the differential equation that satisfies the given initial condition. dy/dx = x/y, y(0) = -2
Using the variable separable method.
Step 1: Separate the variables
dy/dx = x/y
x dx = y dy
Step 2: Integrate both sides of the resulting equation
![\int\limits {x} \, dx = \int\limits{y} \, dy\\(x^2)/(2) = (y^2)/(2) \\ (y^2)/(2) = (x^2)/(2) + C\\y^2 = x^2 + 2C\\y^2 = x^2 + K; K = 2C](https://img.qammunity.org/2021/formulas/mathematics/college/uh5vxndht3njr4e041f8pzij1me9ab8u8a.png)
Note that the constant of integration added to the side containing x
Step 3: Apply the initial condition y(0) = -2
This means when x = 0, y = -2. From step 2:
![y^2+x^2 = K\\(-2)^2+0^2 = K\\4 = K](https://img.qammunity.org/2021/formulas/mathematics/college/gik6d2bqqnganfwmlmc5dgqylyz0ezcp0k.png)
Step 4: Substitute K = 4 into the resulting differential equation above
![y^2=x^2+4\\y = √(x^2+4)](https://img.qammunity.org/2021/formulas/mathematics/college/pil6xlc63uimfkk1hbbr9uclm77ztj4lpx.png)
This gives the solution to the differential equation.