Solve the first order linear inhomogeneous differential equation using the bernoulli method
xy,-2y=2x4
y′−2yx=2x3y' - 2\frac{y}{x} = 2{x^3}y′−2xy=2x3
Substitution: y=uv⇒y′=u′v+uv′y = uv \Rightarrow y' = u'v + uv'y=uv⇒y′=u′v+uv′
Then
u′v+uv′−2uvx=2x3u'v + uv' -2 \frac{{uv}}{x} = 2{x^3}u′v+uv′−2xuv=2x3
u′v+u(v′−2vx)=2x3u'v + u\left( {v' - 2\frac{v}{x}} \right) = 2{x^3}u′v+u(v′−2xv)=2x3
Let
v′−2vx=0⇒dvdx=2vx⇒dvv=2dxx⇒lnv=lnx2⇒v=x2v' - \frac{{2v}}{x} = 0 \Rightarrow \frac{{dv}}{{dx}} = \frac{{2v}}{x} \Rightarrow \frac{{dv}}{v} = \frac{{2dx}}{x} \Rightarrow \ln v = \ln {x^2} \Rightarrow v = {x^2}v′−x2v=0⇒dxdv=x2v⇒vdv=x2dx⇒lnv=lnx2⇒v=x2
u′x2=2x3⇒u′=2x⇒u=x2+C⇒y=uv=(x2+C)x2=x4+Cx2u'{x^2} = 2{x^3} \Rightarrow u' = 2x \Rightarrow u = {x^2} + C \Rightarrow y = uv = \left( {{x^2} + C} \right){x^2} = {x^4} + C{x^2}u′x2=2x3⇒u′=2x⇒u=x2+C⇒y=uv=(x2+C)x2=x4+Cx2
Answer: y=x4+Cx2y = {x^4} + C{x^2}y=x4+Cx2
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