How do I solve the following differential equation:2xy+1+(x2+2y)y′=0

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wyjadaczeqa8 · Accepted Answer

Step 1Note that2xy+x2y′+2yy′=ddx(x2y+y2).So the equation is equivalent to:ddx(x2y+y2)=−1⇒x2y+y2=−x+C⇒y=−x2±x4−4x+4C2y(1)=−1±4C−32=−1⇒C=1Since at x=1 the ± is −, and y is differentiable at any point, the ± will always be − at any point.∴y=−x2−x4−4x+42

calcinaretnbz · Accepted Answer

Step 1Indeed, you can check that I(x,y)=2xy+1 and J(x,y)=x2+2y are both continuously differentiable everywhere and∂I∂y=2x=∂J∂x.A general solution to an exact differential equation is F(x, y)=C for constants C (i.e., level curves of F(x, y)), where F(x,y) is a potential function satisfying∂F∂x=I(x,y) and ∂F∂y=J(x,y). Integrating I(x, y) with respect to x, we obtainF(x,y)=x2y+x+m(y).Integrating J(x,y) with respect to y, we obtainF(x,y)=x2y+y2+n(x).Equating the two expressions for F(x,y), we see thatx2y+x+m(y)=x2y+y2+n(x)⇒m(y)=y2  m{ and }   n(x) = xm{and}  n(x)=x.Thus, a general solution to your differential equation isF(x,y)=x2y+x+y2=C.To find C, we impose the given initial condition y(1)=−1:−1+1+1=1=C.Thus, an implicit solution to your exact differential equation isx2y+x+y2=1.

How do I solve the following differential equation:

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