Answer to Question #157756 in Differential Equations for Jayshree

Question

Solve the following linear differential equations

1) dy/dx + 2y = sinx

2) dy/dx + y = e^x

3) dy/dx +(4/x)y = x³y³

Answer 1

1) $\frac{dy}{dx}+2y=\sin \left(x\right)$  - First order linear Ordinary Differential Equation

A first order linear ODE has the form of     $y'\left(x\right)+p\left(x\right)y=q\left(x\right)$  

Substitute   $\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:+2y=\sin \left(x\right)$

The equation is in the first order linear ODE form

We find the integration factor:  $\mu(x) = e^{2x}$

We put the equation in the form $(\mu(x)y)' = \mu(x)q(x):$     $(e^{2x}y)' = sin(x)e^{2x}$

Solve $(e^{2x}y)' = sin(x)e^{2x} :$            $y=-\frac{\cos \left(x\right)}{5}+\frac{2\sin \left(x\right)}{5}+\frac{c_1}{e^{2x}}$

Answer: $y=\large-\frac{\cos \left(x\right)}{5}+\frac{2\sin \left(x\right)}{5}+\frac{c_1}{e^{2x}}$

2) $\frac{dy}{dx}+y=e^x$ -  First order linear Ordinary Differential Equation

A first order linear ODE has the form of    $y'\left(x\right)+p\left(x\right)y=q\left(x\right)$

Substitute  $\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$\frac{dy}{dx}+y=e^x$

The equation is in the first order linear ODE form

We find the integration factor: $\mu(x) = e^{2x}$

We put the equation in the form $(\mu(x)y)' = \mu(x)q(x):$   $(e^xy)'=e^{2x}$

Solve $(e^xy)'=e^{2x}:$   $y=\frac{e^x}{2}+\frac{c_1}{e^x}$

Answer: $y=\large\frac{e^x}{2}+\frac{c_1}{e^x}$

3) $\frac{dy}{dx}+\left(\frac{4}{x}\right)y=x^3y^3$ -   First order Bernoulli Ordinary Differential Equation

A first order Bernoulli ODE has the form of    $y'+p\left(x\right)y=q\left(x\right)y^n$

Substitute  $\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y' + \left(\frac{4}{x}\right)y=x^3y^3$

The equation is in first order Bernoulli ODE form

The general solution is obtained by substituting $\nu = y^{1-n}$ and solving $\frac{1}{1-n}\nu' + p(x)\nu=q(x)$

Transform to  $\frac{1}{1-n}\nu' + p(x)\nu=q(x)$ :    $\large-\frac{\nu'}{2}+\frac{4\nu}{x}=x^3$

Solve $\large-\frac{\nu'}{2}+\frac{4\nu}{x}=x^3$ :   $\nu = \large\frac{x^4}{2}+c_1x^8$

Substitute back $\nu = y^{-2}:$      $y^{-2} = \large\frac{x^4}{2}+c_1x^8$

Isolate y:     $\large y=\sqrt{\frac{2}{x^4\left(1+2c_1x^4\right)}},\:y=-\sqrt{\frac{2}{x^4\left(1+2c_1x^4\right)}}$

Answer:  $\large y=\sqrt{\frac{2}{x^4\left(1+2c_1x^4\right)}},\:y=-\sqrt{\frac{2}{x^4\left(1+2c_1x^4\right)}}$