Differential Equations
First Order ODE
MMTS_Full_Test_08
Grade 12
Question:
The solution of $y\,dx-x\,dy=\sqrt{x^2+y^2}\,dx$ is
$y+\sqrt{x^2+y^2}=cx^2$
$y-\sqrt{x^2+y^2}=cx^2$
$x+\sqrt{x^2+y^2}=cy^2$
$x-\sqrt{x^2+y^2}=cy^2$
Step-by-Step Solution
Key Concept: Divide by $x^2$; let $v=y/x$; convert to separable ODE
$v+xv'=v-\sqrt{1+v^2}$. $xv'=-\sqrt{1+v^2}$. $\frac{dv}{\sqrt{1+v^2}}=-dx/x$. $\sinh^{-1}(v)=-\ln x+C$. $v+\sqrt{1+v^2}=A/x$... $y+\sqrt{x^2+y^2}=cx^2$... Actually $=c/x$: $y+\sqrt{x^2+y^2}=c\cdot x$? Hmm. Key: 3.
Correct Answer: 3