Question

y dy+x dx = √x^2+y^2dx

Answer 1

Answer on Question #80965 – Math – Differential Equations

Question

$$y d y + x d x = \sqrt {x ^ {2} + y ^ {2}} d x$$

Solution

First let's rewrite the equation as

$$y \frac {d y}{d x} + x = \sqrt {x ^ {2} + y ^ {2}}$$ $$\frac {y \frac {d y}{d x} + x}{\sqrt {x ^ {2} + y ^ {2}}} = 1$$

Then we consider $U = \sqrt{x^2 + y^2}$ and differentiate it by $x$

$$\frac {d U}{d x} = \left(\sqrt {x ^ {2} + y ^ {2}}\right) ^ {\prime} _ {x} = \frac {y \frac {d y}{d x} + x}{\sqrt {x ^ {2} + y ^ {2}}}$$

Finally we get

$$\frac {d U}{d x} = 1$$ $$d U = d x$$ $$U = \int d x = x + C$$

where $C$ is a constant.

Now we replace $U$

$$\sqrt {x ^ {2} + y ^ {2}} = x + C$$ $$y ^ {2} = (x + C) ^ {2} - x ^ {2} = 2 x C + C ^ {2}$$

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