- #1
TogoPogo
- 14
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The problem states:
"By using the substitution [itex]y=xu[/itex], show that the differential equation [itex]\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}, x>0[/itex] can be reduced to the d.e. [itex]x\frac{du}{dx}=\sqrt{u^{2}+1}[/itex].
Hence, show that if the curve passes through the point [itex](1,0)[/itex], the particular solution is given by [itex]y=\frac{1}{2}(x^{2}-1)[/itex]."
I managed to get the d.e. into the form [itex]x\frac{du}{dx}=\sqrt{u^{2}+1}[/itex] but I have no idea how to integrate [itex]\frac{du}{\sqrt{u^{2}+1}}[/itex]. Wolfram Alpha is giving me some inverse hyperbolic sine stuff which I haven't learned yet (I'm in high school). All I've really 'learned' from my teacher so far was solving separable DE's, and inseparable DE's with [itex]y=ux[/itex], however some of the questions that we were given required other techniques like integrating factors and stuff. Is this DE a special case or something?
Anyways, how would I approach this? Do I square both sides to get rid of the square root sign?
Many thanks.
"By using the substitution [itex]y=xu[/itex], show that the differential equation [itex]\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}, x>0[/itex] can be reduced to the d.e. [itex]x\frac{du}{dx}=\sqrt{u^{2}+1}[/itex].
Hence, show that if the curve passes through the point [itex](1,0)[/itex], the particular solution is given by [itex]y=\frac{1}{2}(x^{2}-1)[/itex]."
I managed to get the d.e. into the form [itex]x\frac{du}{dx}=\sqrt{u^{2}+1}[/itex] but I have no idea how to integrate [itex]\frac{du}{\sqrt{u^{2}+1}}[/itex]. Wolfram Alpha is giving me some inverse hyperbolic sine stuff which I haven't learned yet (I'm in high school). All I've really 'learned' from my teacher so far was solving separable DE's, and inseparable DE's with [itex]y=ux[/itex], however some of the questions that we were given required other techniques like integrating factors and stuff. Is this DE a special case or something?
Anyways, how would I approach this? Do I square both sides to get rid of the square root sign?
Many thanks.
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