## The general solution of a differential equation is solving y as a function of x or..y = f(x) Separate both variables using multiplication.\frac{dy}{dx} * ydx = \frac{x}{y} * ydx \\y*dy = x*dxNow integrate both sides in order to find the solution.\int y\, dy = \int x\, dx \\y^2 = x^2 + C Finally, Simplify and solve for y. Remember the constant $\mathcal{C}$ is needed for indefinite integrals!y = \sqrt{x^2 + C}

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Differential Equations

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{'transcript': "Hey, it's Claressa when you mind here. So we're gonna make the equation X D. Why already? Ex plus two, Why was equal to one minus one over X into the form Deal I over the X plus p of X. Why is equal to queue of X? So we know X is bigger than zero. So we divide both equations by X to get d Y over the ex plus two Why over X is equal to one over X minus one over x square. Then we get P of X is equal to her ex and cue of X is equal to one over X minus one over x squared. We see that the integrating factor this sequel to E P E X T X, which is equal to e to the integration of two over X t x, which is equal to eat the l and of X square, which is equal to x square. So we see that when we multiply both sides of the equation This equation by the of X, which is this X square Leah X square you I over d X plus two x y is equal to X minus one So X square. The Why already? Ex plus two x Y is he go to the derivative of X square Y, and this is equal to X minus one. So we're gonna integrate both sides in respect to x tender girl from X minus one D x to get X square plus X square. Why is equal to X square over too minus X plus C, which is equal to why is equal to 1/2 minus one over EPPS plus C over X Square."}

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Differential Equations

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