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4

Hint: Rewrite the DE as $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method

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$$\frac{du}{u}=-\ln(u(t))dt$$ You cannot integrate $\int-\ln(u(t))dt$ directly. The mistake is : You integrate $\quad \int-\ln(u(t))du= -u\ln(u)+u\quad$ but this is not $\quad \int -\ln(u(t))dt$. Back to the original ODE : $$\frac{du}{dt}=u\ln(u)$$ This is a separable ODE thus $$\frac{du}{u\ln(u)}=dt$$ $$\int\frac{du}{u\ln(u)}=\int dt$$ $$\ln(\ln(u))=t+c$$ $$... 1 If we write it as$$ \frac{dy}{dx}-y=-x \, , $$then we are dealing with a linear first-order differential equation of the form$$ \frac{dy}{dx}+p(x)y=q(x) \, . $$To solve an equation of this form, you have to multiply both sides by e^{P(x)}, where P is an antiderivative of p. Here p(x)=-1, and so we can take P(x)=-x. The equation becomes$$ e^{-x}...

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You were on the right track: Let $x_0\in[0,1]$. Then $$f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac 12 f''(c)(x - x_0)^2$$ for $c$ between $x$ and $x_0$. Since $f(x) = 0$ for $x=0$ and $x=1$, it holds that $$0 = f(0) = f(x_0) + f'(x_0)(0 - x_0) + \frac 12 f''(c_1)(0 - x_0)^2$$ and $$0 = f(1) = f(x_0) + f'(x_0)(1 - x_0) + \frac 12 f''(c_2)(1 - x_0)^2.$$ Thus, ...

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Your expression is correct. The key is that "$d\neq \partial$". That is to say, $\frac{d}{dx} f(x,g(x))$ means the derivative of the one-variable function $f(x,g(x))$, whereas $\frac{\partial}{\partial x} f$ means the derivative of the two-variable function $f(x,y)$ with respect to its first argument. It can be helpful to introduce extra symbols ...

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$${y}'\left ( x+ y^{2} \right )= y$$ $$xy'-y=-y'y^2$$ $$\left(\dfrac yx\right)'=-y'\dfrac {y^2}{x^2}$$ This DE is separable. $$\dfrac {x^2}{y^2}d\left(\dfrac yx\right)=-dy$$ Integrate.

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Line (2) is application of the recurrence for $I$. In line (3), write down the first few terms of the first series to help you see what is going on: $$\frac{I(0)t^0}{0!} + \frac{I(1)t^1}{1!} + \frac{I(2)t^2}{2!} + \cdots \text{,}$$ which is $\theta(t)$ To get to line (4), pull out one of the factors to $t$ so the argument to $I$, the power of $t$, and ...

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