Methods of solving first order first degree differential equations

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The slope of a curve at any point on it is the reciprocal of twice the ordinate at that point. If the curve passes through (4,3) then its equation is $lo g(y/3)= \frac{x}{2} -k$ then k is

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Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{y}{2} \Rightarrow \int \frac{d y}{y} \Rightarrow \int \frac{d x}{2} \Rightarrow \log y = \frac{x}{2} + c \\ Passes through (4, 3) then \\ \log 3 = \frac{4}{2} + c \Rightarrow \log 3 = 2 + c \\ \Rightarrow c = \log 3 - 2 \\ \to \log y = \log 3 - 2 + \frac{x}{2} \\ \Rightarrow \log (y / 3) = \frac{x}{2} - 2 \end{array}$

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Methods of solving first order first degree differential equations

Remember concepts with our Masterclasses.

The slope of a curve at any point on it is the reciprocal of twice the ordinate at that point. If the curve passes through (4,3) then its equation is log(y/3)=x2-k then k is

dydx=y2⇒∫dyy⇒∫dx2⇒log⁡y=x2+c Passes through (4,3) then log⁡3=42+c⇒log3=2+c⇒c=log⁡3−2→log⁡y=log⁡3−2+x2⇒log⁡(y/3)=x2−2

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The slope of a curve at any point on it is the reciprocal of twice the ordinate at that point. If the curve passes through (4,3) then its equation is $lo g(y/3)= \frac{x}{2} -k$ then k is