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Solve the following differential equation:`sqrt(1+x^2+y^2+x^2 y^2) +x y(dy)/(dx)=0` |
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Answer» Note that `f(x,y) = sqrt(1+x^(2)+y^(2)+x^(2)y^(2)) = sqrt((1+x^(2))(1+y^(2)))` `sqrt(1+x^(2)) . Sqrt(1+y^(2)) = g(x)h(y)` Thus, f(x,y) can be written as a product of two functions, one of x alone and the othr of y alone. `sqrt(1+x^(2)+y^(2)+x^(2)y^(2))+ xy(dy)/(dx) = 0` `rArr sqrt(1+ x^(2))sqrt(1+y^(2)) + xy (dy)/(dx) = 0` `rArr (sqrt(1+x^(2)))/(x) dx + (y)/(sqrt(1+y^(2)))dy = 0` Integration yields `int (1+x^(2))/(xsqrt(1+x^(2)))dx+(1)/(2)int (d(1+y^(2)))/(sqrt(1+y^(2))) = k` ` rArr int (1)/(xsqrt(1+ x^(2)))dx+ int (x)/(sqrt(1+x^(2))) dx + (1)/(2) xx 2sqrt(1+y^(2)) = k` `rArr int (-(1)/(2)dt)/(((1)/(2))sqrt(1+((1)/(2))^(2)))+ sqrt(1+ x^(2)) + sqrt(1+y^(2)) = k` [setting x = 1/t in the first integral]. `rArr - int (dt)/(sqrt(1+ t^(2))) + sqrt(1+x^(2)) + sqrt(1+y^(2)) = k` `rarr -ln|t + sqrt(1+t^(2))| + sqrt(1+ x^(2)) + sqrt(1+ y^(2)) = k` `rArr -ln |(1)/(x)+sqrt(1+(1)/(x^(2)))| + sqrt(1+x^(2)) + sqrt(1+y^(2)) = k` `rArr sqrt(1+ x^(2)) - ln|(x+sqrt(1+x^(2)))/(x)| + sqrt(1+ y^(2)) = k` which is the general solution of the differential equation. |
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