hello i need help trying to solve this problem. thank you dy/dx=y+xy^3
J josh7 New member Joined Apr 21, 2009 Messages 1 Apr 21, 2009 #1 hello i need help trying to solve this problem. thank you dy/dx=y+xy^3
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Apr 21, 2009 #2 Re: help solving this problem dydx−y=xy3\displaystyle \frac{dy}{dx}-y=xy^{3}dxdy−y=xy3 This is a Bernoulli equation. Let w=y−2, y=w−12, dydx=−12w−32⋅dwdx.......chain rule\displaystyle w=y^{-2}, \;\ y=w^{\frac{-1}{2}}, \;\ \frac{dy}{dx}=-\frac{1}{2}w^{\frac{-3}{2}}\cdot\frac{dw}{dx}.......\text{chain rule}w=y−2, y=w2−1, dxdy=−21w2−3⋅dxdw.......chain rule Make the subs into the original: −12w−32⋅dwdx−w−12=xw−32\displaystyle \frac{-1}{2}w^{\frac{-3}{2}}\cdot\frac{dw}{dx}-w^{\frac{-1}{2}}=xw^{\frac{-3}{2}}2−1w2−3⋅dxdw−w2−1=xw2−3 −12⋅dwdx−w=x\displaystyle \frac{-1}{2}\cdot\frac{dw}{dx}-w=x2−1⋅dxdw−w=x Now, we can use a integrating factor since we transformed it. The IC is e2x\displaystyle e^{2x}e2x ddx[we2x]=−2xe2x\displaystyle \frac{d}{dx}[we^{2x}]=-2xe^{2x}dxd[we2x]=−2xe2x Integrate: we2x=12e2x−xe2x+C\displaystyle we^{2x}=\frac{1}{2}e^{2x}-xe^{2x}+Cwe2x=21e2x−xe2x+C Divide by e^(2x): w=12−x+Ce−2x\displaystyle w=\frac{1}{2}-x+Ce^{-2x}w=21−x+Ce−2x Don't forget to resub w: 1y2=12−x+Ce−2x\displaystyle \frac{1}{y^{2}}=\frac{1}{2}-x+Ce^{-2x}y21=21−x+Ce−2x Try solving for y if you want to whittle it down further. There is a nice stepped through example of a Bernoulli. Now, keep this as a template on future ones. Okey-doke?
Re: help solving this problem dydx−y=xy3\displaystyle \frac{dy}{dx}-y=xy^{3}dxdy−y=xy3 This is a Bernoulli equation. Let w=y−2, y=w−12, dydx=−12w−32⋅dwdx.......chain rule\displaystyle w=y^{-2}, \;\ y=w^{\frac{-1}{2}}, \;\ \frac{dy}{dx}=-\frac{1}{2}w^{\frac{-3}{2}}\cdot\frac{dw}{dx}.......\text{chain rule}w=y−2, y=w2−1, dxdy=−21w2−3⋅dxdw.......chain rule Make the subs into the original: −12w−32⋅dwdx−w−12=xw−32\displaystyle \frac{-1}{2}w^{\frac{-3}{2}}\cdot\frac{dw}{dx}-w^{\frac{-1}{2}}=xw^{\frac{-3}{2}}2−1w2−3⋅dxdw−w2−1=xw2−3 −12⋅dwdx−w=x\displaystyle \frac{-1}{2}\cdot\frac{dw}{dx}-w=x2−1⋅dxdw−w=x Now, we can use a integrating factor since we transformed it. The IC is e2x\displaystyle e^{2x}e2x ddx[we2x]=−2xe2x\displaystyle \frac{d}{dx}[we^{2x}]=-2xe^{2x}dxd[we2x]=−2xe2x Integrate: we2x=12e2x−xe2x+C\displaystyle we^{2x}=\frac{1}{2}e^{2x}-xe^{2x}+Cwe2x=21e2x−xe2x+C Divide by e^(2x): w=12−x+Ce−2x\displaystyle w=\frac{1}{2}-x+Ce^{-2x}w=21−x+Ce−2x Don't forget to resub w: 1y2=12−x+Ce−2x\displaystyle \frac{1}{y^{2}}=\frac{1}{2}-x+Ce^{-2x}y21=21−x+Ce−2x Try solving for y if you want to whittle it down further. There is a nice stepped through example of a Bernoulli. Now, keep this as a template on future ones. Okey-doke?
