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} * y^2dx = \frac{3x^2}{y^2} * y^2dx \\y^2*dy = 3x^2*dxNow integrate both sides in order to find the solution.\int y^2\, dy = \int 3x^2\, dx \\\frac{y^3}{3} = x^3 + C Finally, Simplify and solve for y. Remember the constant $\mathcal{C}$ is needed for indefinite integrals!\frac{y^3}{3}*3 = 3*(x^3+\mathcal{C}) \\y^3 =3x^3+ \mathcal{C} \\y = \sqrt[3]{3x^3 + \mathcal{C}}

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Kayleah T.

Harvey Mudd College

Michael J.

Idaho State University

Boston College

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{'transcript': "he is Clearasil, wouldn't you? Right here. So you have to Why? The derivative is equal to eat the X over too. We put this into plus why we put this into standard for him to get the derivative of y minus 1/2 of y is equal to e to the X over to over two. So we know P of X is equal to negative 1/2 and cue of X is equal to e to the X over too over too. So we get integrating factor which we make as you have X to me. Cool to me to the integration of negative 1/2 D eggs which is equal to eat some negative acts for her too. We multiply both sides by e to the negative X over two and we integrate. She eats the negative X over too. Why derivative plus e to the negative X over too over two times alive is equal to e to the negative x over too times e to the X over too poor too. Then we get Do you ever d x of e to the negative x over too. Why is equal to 1/2 and why is equal to X E to the X over too over two plus c e to the X over too."}

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Differential Equations

GH

Kayleah T.

Harvey Mudd College

Michael J.

Idaho State University

Boston College

Lectures

Join Bootcamp