Question

Question: In Problems 1-30, solve the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{3y}{x}\mathbf{=}{\mathbf{x}}^2\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}$

Short answer

The solution of the given equation is $\mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^3}{6}\mathbf{-}\frac{\mathbf{4}{\mathbf{x}}^2}{5}\mathbf{+}\frac{3x}{4}\mathbf{+}{\mathbf{Cx}}^{-3}$.

Step-by-step solution

Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\frac{3y}{x}\mathbf{=}{\mathbf{x}}^2\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let $\mathbf{P}\left(x\right)\mathbf{=}\frac{3}{x}$.

Find the value of $\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\int \frac{3}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{3}\mathbf{ln}\left|x\right|}\\ \mathbf{=}{\mathbf{x}}^3\end{array}$

Multiply x³ in equation (1) on both sides.

$\begin{array}{c}{\mathbf{x}}^3\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}{\mathbf{x}}^3\frac{3y}{x}\mathbf{=}{\mathbf{x}}^3\left({\mathbf{x}}^2\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\right)\\ {\mathbf{x}}^3\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{3}{\mathbf{x}}^2\mathbf{y}\mathbf{=}{\mathbf{x}}^5\mathbf{-}\mathbf{4}{\mathbf{x}}^4\mathbf{+}\mathbf{3}{\mathbf{x}}^3\\ \frac{d}{\mathrm{dx}}\left[{\mathbf{x}}^3\mathbf{y}\right]\mathbf{=}{\mathbf{x}}^5\mathbf{-}\mathbf{4}{\mathbf{x}}^4\mathbf{+}\mathbf{3}{\mathbf{x}}^3\end{array}$

Integration

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[{\mathbf{x}}^3\mathbf{y}\right]\mathbf{dx}\mathbf{=}\int \left({\mathbf{x}}^5\mathbf{-}\mathbf{4}{\mathbf{x}}^4\mathbf{+}\mathbf{3}{\mathbf{x}}^3\right)\mathbf{dx}\\ {\mathbf{x}}^3\mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^6}{6}\mathbf{-}\frac{\mathbf{4}{\mathbf{x}}^5}{5}\mathbf{+}\frac{\mathbf{3}{\mathbf{x}}^4}{4}\mathbf{+}\mathbf{C}\\ \mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^{-3}{\mathbf{x}}^6}{6}\mathbf{-}\frac{\mathbf{4}{\mathbf{x}}^{-3}{\mathbf{x}}^5}{5}\mathbf{+}\frac{\mathbf{3}{\mathbf{x}}^{-3}{\mathbf{x}}^4}{4}\mathbf{+}{\mathbf{Cx}}^{-3}\\ \mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^3}{6}\mathbf{-}\frac{\mathbf{4}{\mathbf{x}}^2}{5}\mathbf{+}\frac{3x}{4}\mathbf{+}{\mathbf{Cx}}^{-3}\end{array}$

So, the solution is $\mathbf{y}\mathbf{=}\frac{{\mathbf{x}}^3}{6}\mathbf{-}\frac{\mathbf{4}{\mathbf{x}}^2}{5}\mathbf{+}\frac{3x}{4}\mathbf{+}{\mathbf{Cx}}^{-3}$