The solution of the differential equation ydx | vedclass

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(B) Given differential equation is $ydx - (x + 2y^2)dy = 0$.Rearranging the terms,we get $ydx - xdy = 2y^2 dy$.Dividing both sides by $y^2$ (for $y

Vedclass · Accepted Answer

$3$

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(B) Given differential equation is $ydx - (x + 2y^2)dy = 0$.Rearranging the terms,we get $ydx - xdy = 2y^2 dy$.Dividing both sides by $y^2$ (for $y 
eq 0$),we get $\frac{ydx - xdy}{y^2} = 2dy$.This is equivalent to $d(\frac{x}{y}) = 2dy$.Integrating both sides,we get $\frac{x}{y} = 2y + c$.Given $f(-1) = 1$,which means $x = 1$ when $y = -1$. Substituting these values: $\frac{1}{-1} = 2(-1) + c \Rightarrow -1 = -2 + c \Rightarrow c = 1$.Thus,the solution is $\frac{x}{y} = 2y + 1$,or $x = 2y^2 + y$.To find $f(1)$,we substitute $y = 1$ into the equation: $x = 2(1)^2 + 1 = 2 + 1 = 3$.Therefore,$f(1) = 3$.

The solution of the differential equation $ydx - (x + 2y^2)dy = 0$ is $x = f(y)$. If $f(-1) = 1$,then $f(1)$ is equal to

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