Find the equation of the curve passing throug | vedclass

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Question

(A) The given differential equation is $x dy = (2x^2 + 1) dx$.Dividing both sides by $x$ (since $x 
eq 0$),we get $dy = (2x + \frac{1}{x}) dx$.Integr

Vedclass · Accepted Answer

$y = x^2 + \log |x|$

Vedclass · Answer

(A) The given differential equation is $x dy = (2x^2 + 1) dx$.Dividing both sides by $x$ (since $x 
eq 0$),we get $dy = (2x + \frac{1}{x}) dx$.Integrating both sides,we have $\int dy = \int (2x + \frac{1}{x}) dx$.This gives $y = x^2 + \log |x| + C$.Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:$1 = (1)^2 + \log |1| + C$.$1 = 1 + 0 + C$,which implies $C = 0$.Substituting $C = 0$ back into the general solution,we get the required equation: $y = x^2 + \log |x|$.

Find the equation of the curve passing through the point $(1, 1)$ whose differential equation is $x dy = (2x^2 + 1) dx$ where $x \neq 0$.

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