Word Count : 164

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दिया गया अवकल समीकरण [ \frac{dy}{dx}=(x+y)^2 ] मान लें (x+y=z)। तब [ \frac{dz}{dx}=1+\frac{dy}{dx}=1+z^2 ] अतः [ \frac{dz}{1+z^2}=dx ] समाकलन करने पर [ \tan^{-1}z=x+C ] जहाँ (C) समाकलन नियतांक है। पुनः (z=x+y) रखने पर [ \tan^{-1}(x+y)=x+C ] यही दिया गया अवकल समीकरण ________ ___ ____ ___ ______ ________ ______ ___ ___ ______ ____ ________.
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