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Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)` |
Answer» Given differential equation is `ye((x)/(y))dx=(xe^((x)/(y))+y^(2))dy` `implies e^((x)/(y))(dx)/(dy)=(x)/(y)e((x)/(y))+yimpliese^((x)/(y))(dx)/(dy)-(x)/(y)e^((x)/(y))=y` ………`(1)` Let `(x)/(y)=vimpliesx=vyimplies(dx)/(dy)=v+y(dv)/(dx)` From equation `(1)`, `e^(v)(v+y(dv)/(dy))-ve^(v)=y` `implies e^(v)y(dv)/(dy)=yimpliese^(v)(dv)/(dy)=1impliese^(v)dv=dy` On integration, `inte^(v)dv=int1dyimpliese^(v)=y+Cimpliese^((x)/(y))=y+C` |
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