The curve satisfying the differential equation, ydx -(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
Solutions · 1
Official solution
Given,
y dx = \[\left( {x + 3{y^2}} \right)dy\]
\[ \Rightarrow \] ydydx= x + 3y2⇒\[\,\,\,\] \[{{dx} \over {dy}}\] - \[{x \over y} = 3y\]
If = \[{e^{ - \int {{1 \over y}dy} }}\] = \[{e^{ - \ln y}}\] = \[{1 \over y}\]
\[\therefore\,\,\,\] Soluation is ,
x . \[{1 \over y}\] = \[\int {3y.{1 \over y}dy} \]
\[ \Rightarrow \] {x \over y}\[ = 3y + c
This curve passing through (1, 1)
\]\therefore\,\,\,1 = 3 + c⇒\[\,\,\,\] c = - 2
\[\therefore\,\,\,\] Curve is, x = 3y2 - 2y
Now put every point in this equation, and see which point satisfy this equation.
Following this method you can see (- \[{1 \over 3}\], \[{1 \over 3}\]) point satisfy this equation.