2yy' +3 = y^2 + 3x with initial value y(0) = 8
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Originally Posted by mmiraj 2yy' +3 = y^2 + 3x with initial value y(0) = 8 One way to start this is to note that , so your equation becomes: Now, is just a function, let's call it , turning your equation into Can you finish from here? -Dan
How is 2yy' = (y^2) ???
Originally Posted by mmiraj How is 2yy' = (y^2) ??? It’s not. topsquark posted $2yy’ = (y^2)’$ ... note the prime symbol after the parentheses, i.e. the derivative of $y^2$ is $2y \cdot y’$
Originally Posted by topsquark One way to start this is to note that , so your equation becomes: Now, is just a function, let's call it , turning your equation into Can you finish from here? -Dan Let $u = y^2+3x$. Now you have $u' = u$.
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