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Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)`A. `e^(x//y)=cosy+C`B. `e^(x//y)=-siny+C`C. `e^(y//x)=-cosy+C`D. `e^(x//y)=-cosy+C` |
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Answer» Correct Answer - C We have, `ye^(x//y)dx=(xe^(x//y)+y^(2)siny)dy` `rArr" "(ydx-xdy)e^(x//y)=y^(2)sinydy` `rArr" "e^(x//y){(ydx-xdy)/(y^(2))}=sin ydy` `rArr" "e^(x//y)d((x)/(y))=-d(cosy)` On integrating, we obtain `e^(x//y)=-cosy+C`, as the required solution. |
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