# Evaluate: (dy)/(dy)+(3y)/(x)=6x^2

## Expression: $\frac{ \mathrm{d}y }{ \mathrm{d}y }+\frac{ 3y }{ x }=6{x}^{2}$

Write the expression as a product with the factor $y$

$\frac{ \mathrm{d}y }{ \mathrm{d}y }+\frac{ 3 }{ x } \times y=6{x}^{2}$

Since the equation is written in standard form, determine the functions $P\left( x \right)$ and $Q\left( x \right)$

$\begin{array} { l }P\left( x \right)=\frac{ 3 }{ x },\\Q\left( x \right)=6{x}^{2}\end{array}$

Insert the function $P\left( x \right)=\frac{ 3 }{ x }$ into the formula for the integrating factor $u\left( x \right)$

$\begin{array} { l }u\left( x \right)={e}^{\int{ \frac{ 3 }{ x } } \mathrm{d} x},\\Q\left( x \right)=6{x}^{2}\end{array}$

Evaluate the integral

$\begin{array} { l }u\left( x \right)={e}^{3\ln\left({x}\right)},\\Q\left( x \right)=6{x}^{2}\end{array}$

Use $x \times \ln\left({a}\right)=\ln\left({{a}^{x}}\right)$ to transform the expression

$\begin{array} { l }u\left( x \right)={e}^{\ln\left({{x}^{3}}\right)},\\Q\left( x \right)=6{x}^{2}\end{array}$

Use ${e}^{\ln\left({x}\right)}=x$ to simplify the expression

$\begin{array} { l }u\left( x \right)={x}^{3},\\Q\left( x \right)=6{x}^{2}\end{array}$

Insert the integrating factor $u\left( x \right)$ and the function $Q\left( x \right)$ into the general solution formula

$y=\frac{ 1 }{ {x}^{3} } \times \int{ 6{x}^{2} \times {x}^{3} } \mathrm{d} x$

Calculate the product

$y=\frac{ 1 }{ {x}^{3} } \times \int{ 6{x}^{5} } \mathrm{d} x$

Evaluate the integral

$\begin{array} { l }y=\frac{ 1 }{ {x}^{3} } \times \left( {x}^{6}+C \right),& C \in ℝ\end{array}$

Multiply each term in the parentheses by $\frac{ 1 }{ {x}^{3} }$

$\begin{array} { l }y=\frac{ 1 }{ {x}^{3} } \times {x}^{6}+\frac{ 1 }{ {x}^{3} } \times C,& C \in ℝ\end{array}$

Cancel out the common factor ${x}^{3}$

$\begin{array} { l }y={x}^{3}+\frac{ 1 }{ {x}^{3} } \times C,& C \in ℝ\end{array}$

Calculate the product

$\begin{array} { l }y={x}^{3}+\frac{ C }{ {x}^{3} },& C \in ℝ\end{array}$

Random Posts
Random Articles