Question ID: #741
Let the solution curve of the differential equation $xdy – ydx = \sqrt{x^{2}+y^{2}}dx,$ $x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to
- (1) 4
- (2) 6
- (3) 1
- (4) 2
Solution:
Rearranging the given differential equation:
$$ xdy – ydx = \sqrt{x^2+y^2} dx $$
$$ \frac{xdy – ydx}{x^2} = \frac{\sqrt{x^2+y^2}}{x^2} dx $$
Recall that $d\left(\frac{y}{x}\right) = \frac{xdy – ydx}{x^2}$.
Also, inside the square root on the RHS, factor out $x^2$:
$$ \frac{\sqrt{x^2(1+\frac{y^2}{x^2})}}{x^2} dx = \frac{x\sqrt{1+(\frac{y}{x})^2}}{x^2} dx = \frac{1}{x}\sqrt{1+\left(\frac{y}{x}\right)^2} dx $$
So the equation becomes:
$$ d\left(\frac{y}{x}\right) = \frac{1}{x} \sqrt{1+\left(\frac{y}{x}\right)^2} dx $$
$$ \frac{d\left(\frac{y}{x}\right)}{\sqrt{1+\left(\frac{y}{x}\right)^2}} = \frac{dx}{x} $$
Integrating both sides:
$$ \int \frac{dt}{\sqrt{1+t^2}} = \int \frac{dx}{x} $$
$$ \ln|t + \sqrt{1+t^2}| = \ln|x| + \ln C $$
$$ t + \sqrt{1+t^2} = Cx $$
Substitute $t = \frac{y}{x}$:
$$ \frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = Cx $$
$$ y + \sqrt{x^2+y^2} = Cx^2 $$
Using the initial condition $y(1) = 0$:
$$ 0 + \sqrt{1^2+0} = C(1)^2 \Rightarrow C = 1 $$
The particular solution is:
$$ y + \sqrt{x^2+y^2} = x^2 $$
We need to find $y(3)$. Let $y(3) = k$.
$$ k + \sqrt{3^2+k^2} = 3^2 $$
$$ \sqrt{9+k^2} = 9 – k $$
Squaring both sides (condition $9-k \ge 0 \Rightarrow k \le 9$):
$$ 9 + k^2 = (9 – k)^2 $$
$$ 9 + k^2 = 81 + k^2 – 18k $$
$$ 9 = 81 – 18k $$
$$ 18k = 72 \Rightarrow k = 4 $$
So, $y(3) = 4$.
Ans. (1)
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