Question

Solve the initial value problem yy′+x=x2+y2‾‾‾‾‾‾‾√ with y(1)=8‾√. To solve this, we should use the substitution u= x^2+y^2 help (formulas) u′= 2x+2yy' help (formulas) Enter derivatives using prime notation (e.g., you would enter y′ for dydx). After the substitution from the previous part, we obtain the following linear differential equation in x,u,u′.

Answer 1

Looks like the equation is

$yy'+x=\sqrt{x^2+y^2}$

Substitute $u(x)=x^2+y(x)^2$, so that $u'=2x+2yy'$. Then the equation is the same as

$\dfrac{u'-2x}2+x=\sqrt u\implies u'=2\sqrt u\implies\dfrac{\mathrm du}{2\sqrt u}=\mathrm dx$

Integrate both sides to get

$\sqrt u=C\implies\sqrt{x^2+y^2}=C$

Given that $y(1)=\sqrt8$, we have

$\sqrt{1^2+(\sqrt8)^2}=C\implies C=3$

so the solution is

$\sqrt{x^2+y^2}=3$