Like
Report
In Exercises $1-14,$ find the general solution of the differential equation.
$$
\frac{d y}{d x}=\frac{x}{y}
$$
The general solution of a differential equation is solving y as a function of x or..
\begin{equation}
y = f(x)
\end{equation}
Separate both variables using multiplication.
\begin{equation}
\frac{dy}{dx} * ydx = \frac{x}{y} * ydx \\
y*dy = x*dx
\end{equation}
Now integrate both sides in order to find the solution.
\begin{equation}
\int y\, dy = \int x\, dx \\
y^2 = x^2 + C
\end{equation}
Finally, Simplify and solve for y. Remember the constant \(\mathcal{C}\) is needed for indefinite integrals!
\begin{equation}
y = \sqrt{x^2 + C}
\end{equation}
Differential Equations
Numerade Educator
Oregon State University
Baylor University
Idaho State University
{'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."}