Question

\(y^{\prime 2}-2 x y^{\prime}-8 x^{2}=0\)

Short answer

The obtained first-order linear differential equation is \(y'(x) - 4x^2y(x) = 2x + C\).

Step-by-step solution

Rewrite the differential equation in terms of \(y'\)

We can rewrite the given first-order differential equation as \(y^{\prime}(y^\prime-2x)=8x^2\). This expression separates the terms containing \(y'\) on the left side and the terms containing \(x\) on the right side.

Divide both sides by \(y'^2\)

Next, we'll divide both sides by \(y'^2\) to fully separate the variables: \[-\frac{y^\prime-2x}{y^{\prime 2}}=-\frac{8x^2}{y^{\prime 2}}\]

Apply a substitution

Let \(u = y' - 2x\). Then, \(y' = u + 2x\), and the differential becomes \(\frac{dy}{dx} = u + 2x\). Rearranging and solving for \(dx\), we get \(dx = \frac{dy}{u+2x}\). Now, substitute the expression for \(u\) back into the expression we obtained in Step 2: \[-\frac{u}{(u+2x)^2} = -\frac{8x^2}{(u+2x)^2}\]

Integrate both sides with respect to \(y\)

Now, we'll integrate both sides with respect to \(y\): \[\int \frac{u}{(u + 2x)^2} dy = \int \frac{8x^2}{(u + 2x)^2} dy\] Apply the substitution we created in Step 3 on both sides of the equation for \(dx\): \[\int \frac{1}{u + 2x} du = \int \frac{8x^2}{(u + 2x)^2} \frac{dy}{u + 2x}\] Cancel out the common \((u + 2x)\) term in both integrals: \[\int du = \int 8x^2 dy\]

Integrate and solve for \(y\)

Integrate both sides of the equation: \[u = 4x^2y + C\] Substitute back \(y'\) for \(u\): \[y' - 2x = 4x^2y + C\] Now, solve the first-order linear differential equation for \(y(x)\): \[y'(x) - 4x^2y(x) = 2x + C\] Solve this inhomogeneous differential equation using the integrating factor method or any other appropriate method to find the unknown function \(y(x)\), which will be the general solution to the given differential equation.