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$$x\frac{dy}{dx}=x^2 +y$$

given that $\\ y\left( 1 \right) =0$

When i got partial derivatives of both sides, found it's not an exact equation..please can anybody can give a clue to solve this..

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$$x\frac { dy }{ dx } =x^{ 2 }+y\\ xdy-ydx={ x }^{ 2 }dx\\ \frac { xdy-ydx }{ { x }^{ 2 } } =dx\\ \\ d\left( \frac { y }{ x } \right) =dx\\ \frac { y }{ x } =x+C\\ y=x\left( x+C \right) \\ $$ and since $y(1)=0$ we get from that $$y\left( x \right) =x\left( x+C \right) \\ y\left( 1 \right) =0\\ 1+C=0\\ C=-1$$ finally

$$y\left( x \right) =x\left( x-1 \right) ={ x }^{ 2 }-x$$

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  • $\begingroup$ @peterwhy,than you $\endgroup$ – haqnatural Jun 20 '17 at 23:22
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The standard way of solving this type of equation would be to notice it is a linear differential equation: $$\frac{dy}{dx}-\frac{1}{x}\cdot y=x$$ So, the integrating factor here is $e^{\int -\frac{1}{x} dx} = \frac{1}{x}$, and we can write $$\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=1$$ $$\frac{d}{dx}\left(\frac{y}{x}\right)=1$$ $$\frac{y}{x}=x+c$$ $$y=x^2+c x$$ Finally, applying the initial condition gives $$y=x^2- x$$

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Two good methods of solution have been given. Here is a third using an integration factor.

Since the initial equation is not exact we can check to see whether either

  1. $\dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}$ is a function of $x$ alone or whether

  2. $-\dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{M}$ is a function of $y$ alone

In this case we see that $\dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}=-\dfrac{2}{x}$

which gives an integrating factor of $\mu=\int e^{-2/x}dx=x^{-2}$.

Applying the integrating factor yields the exact equation

$$ \left(1+x^{-2}y\right)\,dx-x^{-1}dy=0 $$

which can then easily be integrated yielding $y=x^2-x$.

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Consider it as a linear equation in $y$ \begin{eqnarray*} x\frac{dy}{dx}-y= x^2 \end{eqnarray*} First solve \begin{eqnarray*} x\frac{dy}{dx}-y= 0 \end{eqnarray*} Assume the solution $y=x^{\lambda}$ ... $\lambda=1$ and we have the general solution $y=Ax$.

Now for the particular solution assume the form $y=Bx^2$ substitute this back into the original eqation and we have $B=1$. Now put in the initial condtion & $\color{red}{y=x^2-x}$.

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$x\frac{dy}{dx}-y=x^2$. Divide both sides by $x$ to get $y'-\frac{y}{x}=x$, the integrating factor is $e^{\int-\frac{1}{x}\,\mathrm{d}x}=\frac{1}{x}$. Our differential equation then becomes $\left(\frac{y}{x}\right)'=1\implies\frac{y}{x}=x+C\implies\boxed{y=x^2+Cx}$ solving for the constant gives us $C=-1$

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