# How i solve differential equation x dy/dx -y=2x^2 and x dy/dx +y=3 ?

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#### Explanation

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#### Explanation:

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8
Sep 20, 2016

If $x y ' - y = 2 {x}^{2}$, then $y = 2 {x}^{2} + C x$.

#### Explanation:

Start with the first: $x \frac{\mathrm{dy}}{\mathrm{dx}} - y = 2 {x}^{2}$.

Standard form for a linear differential equation is $y ' + p \left(x\right) y = q \left(x\right) .$

Divide by $x$ to put into standard form: $y ' - \frac{1}{x} y = 2 x$.

Now we need to multiply both sides by an integrating factor $I \left(x\right)$.

$I \left(x\right) = {e}^{\int p \left(x\right) \mathrm{dx}}$.

In this case $p \left(x\right) = - \frac{1}{x}$.

$\int p \left(x\right) \mathrm{dx} = \int - \frac{1}{x} \mathrm{dx} = - \ln x = \ln \left(\frac{1}{x}\right)$.

So $I \left(x\right) = {e}^{\ln} \left(\frac{1}{x}\right) = \frac{1}{x}$.

Multiply the entire equation in standard form by $\frac{1}{x}$ to get $\frac{1}{x} y ' - \frac{1}{x} ^ 2 y = 2$.

At this point you should be asking why on earth did we just multiply this by some random function? The reason why is now the left hand side is expressible as the derivative of the product of $y$ and our integrating factor.

In other words, we now have $\frac{d}{\mathrm{dx}} \left[\frac{1}{x} y\right] = 2$. You can differentiate to make sure that $\frac{d}{\mathrm{dx}} \left[\frac{1}{x} y\right]$ actually equals $\frac{1}{x} y ' - \frac{1}{x} ^ 2 y$ as it should.

Now that we have $\frac{d}{\mathrm{dx}} \left[\frac{1}{x} y\right] = 2$, we can integrate both sides to get $\frac{1}{x} y = \int 2 \mathrm{dx}$, so we have $\frac{1}{x} y = 2 x + C$

Finally to solve for $y$, multiply both sides by $x$ to get $y = 2 {x}^{2} + C x$.

Let's make sure that this is true by plugging our answer back into the original differential equation.

$\textcolor{b l u e}{y} = \textcolor{b l u e}{2 {x}^{2} + C x}$.

$\textcolor{red}{y '} = \textcolor{red}{4 x + C}$.

We want to show that $x \textcolor{red}{y '} - \textcolor{b l u e}{y} = 2 {x}^{2}$.

We have $x \left(\textcolor{red}{4 x + C}\right) - \textcolor{b l u e}{\left(2 {x}^{2} + C x\right)} = 4 {x}^{2} + C x - 2 {x}^{2} - C x = 2 {x}^{2}$, which is what we wanted to show, so our solution is correct.

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