Question

In Problems 1–22 solve the given differential equation by separation

of variables.

$\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{4}\mathit{y}$

Short answer

The solution of the given differential equation is.$y=C{x}^4$

Step-by-step solution

Step 1:Separable Equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables and integrate

The given equation is,$x\frac{dy}{dx}=4y$.

Separate the variables,

$\frac{dy}{y}=\frac{4dx}{x}$

Integrate both sides,

$\int \frac{dy}{y}=4\int \frac{dx}{x}$

Apply,$\int \frac{du}{u}=\ln \left|u\right|$so

$\ln \left|y\right|=4\ln \left|x\right|+C_1$

Apply e to both sides,

$\begin{array}{l}{e}^{\ln \left|y\right|}=e^{4\ln \left|x\right|+C_1}\\ e^{\ln \left|y\right|}=e^{4\ln \left|x\right|}\cdot e^{C_1}\end{array}$

Where, $e^{C_1}=C$, so

$e^{\ln \left|y\right|}=C{e}^{4\ln \left|x\right|}$

Apply, $a\ln b=\ln b^4$, so

$e^{\ln \left|y\right|}=C{e}^{\ln x^4}$

Use,$e^{\ln z}=z$so

$y=C{x}^4$

Hence, the solution of the given differential equation is.$y=C{x}^4$