Notes: `y' + 2xy = 10x` Solve the first-order differential equation

Monday, 14 September 2015

$\displaystyle{y}'+{2}{x}{y}={10}{x}$ Solve the first-order differential equation

$\displaystyle{y}'+{2}{x}{y}={10}{x}$

To solve, re-write the derivative as $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ .

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+{2}{x}{y}={10}{x}$

Then, bring together same variables on one side of the equation.

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={10}{x}-{2}{x}{y}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={2}{x}{\left({5}-{y}\right)}$

$\displaystyle\frac{{\left.{d}{y}}}{{{5}-{y}}}={2}{x}{\left.{d}{x}\right.}$

Next, take the integral of both sides.

$\displaystyle\int\frac{{\left.{d}{y}}}{{{5}-{y}}}=\int{2}{x}{\left.{d}{x}\right.}$

$\displaystyle-{\ln}{\left|{5}-{y}\right|}+{C}_{{1}}=\frac{{{2}{{x}}^{{2}}}}{{2}}+{C}_{{2}}$

Then, isolate the y.

$\displaystyle-{\ln}{\left|{5}-{y}\right|}={{x}}^{{2}}+{C}_{{2}}-{C}_{{1}}$

$\displaystyle{\ln}{\left|{5}-{y}\right|}=-{{x}}^{{2}}-{C}_{{2}}+{C}_{{1}}$

Since C1 and C2 represent any number,...

$\displaystyle{y}'+{2}{x}{y}={10}{x}$

To solve, re-write the derivative as $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ .

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}+{2}{x}{y}={10}{x}$

Then, bring together same variables on one side of the equation.

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={10}{x}-{2}{x}{y}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={2}{x}{\left({5}-{y}\right)}$

$\displaystyle\frac{{\left.{d}{y}}}{{{5}-{y}}}={2}{x}{\left.{d}{x}\right.}$

Next, take the integral of both sides.

$\displaystyle\int\frac{{\left.{d}{y}}}{{{5}-{y}}}=\int{2}{x}{\left.{d}{x}\right.}$

$\displaystyle-{\ln}{\left|{5}-{y}\right|}+{C}_{{1}}=\frac{{{2}{{x}}^{{2}}}}{{2}}+{C}_{{2}}$

Then, isolate the y.

$\displaystyle-{\ln}{\left|{5}-{y}\right|}={{x}}^{{2}}+{C}_{{2}}-{C}_{{1}}$

$\displaystyle{\ln}{\left|{5}-{y}\right|}=-{{x}}^{{2}}-{C}_{{2}}+{C}_{{1}}$

Since C1 and C2 represent any number, express it as a single constant C.

$\displaystyle{\ln}{\left|{5}-{y}\right|}=-{{x}}^{{2}}+{C}$

$\displaystyle{{e}}^{{{\ln}{\left|{5}-{y}\right|}}}={{e}}^{{-{{x}}^{{2}}+{C}}}$

$\displaystyle{\left|{5}-{y}\right|}={{e}}^{{-{{x}}^{{2}}+{C}}}$

$\displaystyle{5}-{y}=\pm{{e}}^{{-{{x}}^{{2}}+{C}}}$

Applying the exponent rule $\displaystyle{{a}}^{{m}}\cdot{{a}}^{{n}}={{a}}^{{{m}+{n}}}$ ,

the right side becomes

$\displaystyle{5}-{y}=\pm{{e}}^{{-{{x}}^{{2}}}}\cdot{{e}}^{{C}}$

$\displaystyle{5}-{y}=\pm{{e}}^{{C}}\cdot{{e}}^{{-{{x}}^{{2}}}}$

$\displaystyle-{y}=\pm{{e}}^{{C}}\cdot{{e}}^{{-{{x}}^{{2}}}}-{5}$

$\displaystyle{y}=\pm{{e}}^{{C}}\cdot{{e}}^{{-{{x}}^{{2}}}}+{5}$

Since+-e^C is a constant, it can be replaced by a constant C.

$\displaystyle{y}={C}{{e}}^{{-{{x}}^{{2}}}}+{5}$

Therefore, the general solution is $\displaystyle{y}={C}{{e}}^{{-{{x}}^{{2}}}}+{5}$ .

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