Solve the separable differential equation 8x-3ysqrt(x^2+1)dy/dx=0 subject to initial condition y(0) = 1 This is what I have done so far.... -3ysqrt(x^2+1)dy/dx = -8x -3ydy = -8x/sqrt(x^2+1)dx -3/2y^2 = ?
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Originally Posted by derekjonathon Solve the separable differential equation 8x-3ysqrt(x^2+1)dy/dx=0 subject to initial condition y(0) = 1 This is what I have done so far.... -3ysqrt(x^2+1)dy/dx = -8x -3ydy = -8x/sqrt(x^2+1)dx -3/2y^2 = ? For the right integral, let so . Now it's so...
Originally Posted by derekjonathon Solve the separable differential equation 8x-3ysqrt(x^2+1)dy/dx=0 subject to initial condition y(0) = 1 This is what I have done so far.... -3ysqrt(x^2+1)dy/dx = -8x -3ydy = -8x/sqrt(x^2+1)dx -3/2y^2 = ? where Now you know . Therefore .
Never mind, you fixed it.
Originally Posted by redsoxfan325 I think you lost a square root sign going from step 2 to step 3. Yes I did. Edited.
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