Nonhomogeneous 2nd order dif equation

darjiaus7
Messages
1
Reaction score
0
Hello guys . I want to do nonhomogeneous 2nd order dif equation..I am trying to do this for 2 days , but I can't get good answer. Can you show me how to do this equation with constant variation method ( i know it best) or other I would be very gratefull , because after 2 days of trying I am surrendered... xy''-2y' = -\frac{2}{x^2} ; y(1)=-\frac{1}{3} ; y'(1)=1
 
Physics news on Phys.org
darjiaus7 said:
Hello guys . I want to do nonhomogeneous 2nd order dif equation..I am trying to do this for 2 days , but I can't get good answer. Can you show me how to do this equation with constant variation method ( i know it best) or other I would be very gratefull , because after 2 days of trying I am surrendered... xy''-2y' = -\frac{2}{x^2} ; y(1)=-\frac{1}{3} ; y'(1)=1

This is a first-order ODE for u = y&#039;: <br /> u&#039; -\frac{2}{x}u = -\frac{2}{x^3}. It has an integrating factor.
 
change variables
##\mathrm{u}(x)=x^{-2} \mathrm{y}^\prime (x)##
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top