(2x+1)y′=4x+2yy=u(x)v(x)(2x+1)(u′v+uv′)=4x+2uv(2x+1)u′v+(2x+1)uv′−2uv=4x(2x+1)u′v+u((2x+1)v′−2v)=4x1)(2x+1)v′−2v=0vdv=2x+12dx∫vdv=∫2x+12dxln∣v∣=ln∣2x+1∣v=2x+12)(2x+1)u′⋅(2x+1)=4xdu=(2x+1)24xu=∫(2x+1)24xdx==∫(2x+1)22(2x+1−1)dx==∫(2x+12−(2x+1)22)dx==ln∣2x+1∣+2x+11+Cy=(2x+1)(ln∣2x+1∣+2x+11+C)
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