Ernest Ryland

## Answered question

2021-12-26

Solve the given differential equation. If an initial condition is given, also find the solution that satisfies it.
$\frac{dy}{dx}=3-6x+y-2xy$

### Answer & Explanation

Thomas Nickerson

Beginner2021-12-27Added 32 answers

Step 1
To solve the differential equation
$\frac{dy}{dx}=3-6x+y-2xy$
Step 2
$\frac{dy}{dx}=3-6x+y-2xy$
$\frac{dy}{dx}=3\left(1-2x\right)+y\left(1-2x\right)$
$\frac{dy}{dx}=\left(3+y\right)\left(1-2x\right)$
$\frac{dy}{3+y}=\left(1-2x\right)dx$
Compare with
$P\left(y\right)dy=Q\left(x\right)dx$
$P\left(y\right)=\frac{1}{3+y}$
$Q\left(x\right)=1-2x$
Thus, given differential equation can be solved using variable separable type of differential equation.
Step 3
$\frac{dy}{3+y}=\left(1-2x\right)dx$
Integrate both sides
$\int \frac{dy}{3+y}=\int \left(1-2x\right)dx$
$\mathrm{ln}\left(3+y\right)=x-{x}^{2}+\mathrm{ln}\left(C\right)$
$\mathrm{ln}\left(3+y\right)-\mathrm{ln}\left(C\right)=x-{x}^{2}$
$\mathrm{ln}\left(\frac{3+y}{C}\right)=x-{x}^{2}$
$\frac{3+y}{C}={e}^{x-x2}$
$3+y=C{e}^{x-x2}$
$y=C{e}^{x-x2}-3$

Cleveland Walters

Beginner2021-12-28Added 40 answers

${y}^{\prime }=3-6x+y-2xy$
$\frac{1}{y+3}{y}^{\prime }=1-2x$
Solve $\frac{1}{y+3}{y}^{\prime }=1-2x:\mathrm{ln}\left(y+3\right)=x-{x}^{2}+{c}_{1}$
Isolate y: $y={e}^{x-{x}^{2}+{c}_{1}}-3$

karton

Expert2022-01-09Added 613 answers

This is separable, after some factoring. We have
$\frac{dy}{dx}=3-6x+y-2xy$
$=3\left(1-2x\right)+y\left(1-2x\right)$
$=\left(y+3\right)\left(1-2x\right)$
which after separating becomes

Integrate both sides, we get $\mathrm{ln}|y+3|=x-{x}^{2}+C$
both sides, we get $y=-3+c{e}^{x-{x}^{2}}$
This is also linear. A good practice exercise would be to solve it as linear ODE and make the solutions match.

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