Deduce that for $f \in C^{2}(\mathbf{R}) \cap CL(\mathbf{R})$ such that $f^{\prime}$ and $f^{\prime\prime}$ have slow growth, we have
$$\forall t \in \mathbf{R}_{+}^{*}, \forall x \in \mathbf{R}, \quad \frac{\partial P_{t}(f)(x)}{\partial t} = L\left(P_{t}(f)\right)(x)$$
where $\forall x \in \mathbf{R}, \quad L(f)(x) = f^{\prime\prime}(x) - x f^{\prime}(x).$

We admit that $S$ is of class $C^{1}$ on $\mathbf{R}_{+}^{*}$ and that
$$\forall t \in \mathbf{R}_{+}^{*}, \quad S^{\prime}(t) = \int_{-\infty}^{+\infty} \frac{\partial P_{t}(f)(x)}{\partial t} \left(1 + \ln\left(P_{t}(f)(x)\right)\right) \varphi(x) \mathrm{d}x$$
Show that
$$\forall t \in \mathbf{R}_{+}^{*}, \quad S^{\prime}(t) = \int_{-\infty}^{+\infty} L\left(P_{t}(f)\right)(x) \left(1 + \ln\left(P_{t}(f)(x)\right)\right) \varphi(x) \mathrm{d}x$$

Deduce that for $f \in C^2(\mathbf{R}) \cap CL(\mathbf{R})$ such that $f'$ and $f''$ have slow growth, we have $$\forall t \in \mathbf{R}_+^*, \forall x \in \mathbf{R}, \quad \frac{\partial P_t(f)(x)}{\partial t} = L\!\left(P_t(f)\right)(x),$$ where $L(f)(x) = f''(x) - xf'(x)$.

We admit that $S$ is of class $C^1$ on $\mathbf{R}_+^*$ and that $$\forall t \in \mathbf{R}_+^*, \quad S'(t) = \int_{-\infty}^{+\infty} \frac{\partial P_t(f)(x)}{\partial t}\left(1 + \ln\!\left(P_t(f)(x)\right)\right)\varphi(x)\,\mathrm{d}x.$$ Show that $$\forall t \in \mathbf{R}_+^*, \quad S'(t) = \int_{-\infty}^{+\infty} L\!\left(P_t(f)\right)(x)\left(1 + \ln\!\left(P_t(f)(x)\right)\right)\varphi(x)\,\mathrm{d}x.$$

Show that a power series $f(x) = \sum_{n=0}^{\infty} \frac{c_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if its Laplace transform $$\widehat{f}(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} c_n x^n$$ is a solution of a differential equation.

Using equation $(E)$ satisfied by $y$, calculate $b _ { 1 }$.
The equation $(E)$ is $y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }$, and $y$ is the sum of a Dirichlet series $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ with $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.

Let $k \in \mathbf { N } ^ { * }$. Using equation $(E)$ satisfied by $y$, exhibit a recurrence relation linking $b _ { k + 1 } , b _ { k }$ and $d _ { k }$.
The equation $(E)$ is $y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }$, with $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$ and $d_k$ as defined in question 11.

Suppose that $S _ { 0 } > 0$. Show that the function $S$ of the solution triplet $( S , I , R )$ of $(F)$ satisfies the relation
$$\left( - \frac { S ^ { \prime } } { S } \right) ^ { \prime } = - S ^ { \prime } + \frac { S ^ { \prime } } { S }$$
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$

We suppose in this question that $f \in \mathcal{C}^1(\mathbb{R})$ is convex, and that $f'$ is $L$-Lipschitzian, for some $L > 0$. a) Show that for all $x, y \in \mathbb{R}$ $$\left|f'(x) - f'(y)\right|^2 \leq L(x-y)\left(f'(x) - f'(y)\right)$$ b) Let $x, y \in \mathbb{R}$, and let $\tilde{x} := x - \tau f'(x)$ and $\tilde{y} := y - \tau f'(y)$. Show that $$|\tilde{x} - \tilde{y}|^2 \leq |x-y|^2 - \tau(2 - \tau L)(x-y)\left(f'(x) - f'(y)\right)$$ c) We further suppose that $f$ admits a minimizer $x_*$, and that $0 < \tau \leq 2/L$. Show that the sequence $\left(\left|x_n - x_*\right|\right)_{n \in \mathbb{N}}$ is decreasing. (Recall that $\left(x_n\right)_{n \in \mathbb{N}}$ satisfies the recurrence relation $\forall n \in \mathbb{N},\, x_{n+1} := x_n - \tau f'(x_n)$.)

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and we fix $\tau$ such that $0 < \tau \leq 2/L$. We further suppose that $f$ is $\alpha$-convex, with $\alpha > 0$, that is $$g(x) := f(x) - \frac{1}{2}\alpha x^2 \quad \text{is a convex function on } \mathbb{R}$$ Justify that $f'(x) - \alpha x$ is an increasing function of $x \in \mathbb{R}$. Deduce that $\alpha \leq L$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and $f$ is $\alpha$-convex with $\alpha > 0$, that is $g(x) := f(x) - \frac{1}{2}\alpha x^2$ is a convex function on $\mathbb{R}$. Show that $f(x) \geq f(0) + f'(0)x + \alpha x^2/2$ for all $x \in \mathbb{R}$. Deduce that $f$ admits a minimizer on $\mathbb{R}$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, $f$ is $\alpha$-convex with $\alpha > 0$, and $x_*$ denotes a minimizer of $f$. Show that for all $x, y \in \mathbb{R}$ $$\alpha|x-y|^2 \leq \left(f'(x) - f'(y)\right)(x-y)$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we suppose that $f \in \mathcal{C}^1(\mathbb{R})$. Show that $x_1 := p_f(x_0)$ satisfies $$x_1 = x_0 - \tau f'(x_1)$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$, where $p_f(x_0)$ is the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we set $f(x) := |x|$ for all $x \in \mathbb{R}$. a) Show that $$p_f(x) = \begin{cases} x - \tau & \text{if } x \geq \tau \\ x + \tau & \text{if } x \leq -\tau \\ 0 & \text{otherwise} \end{cases}$$ b) Deduce that $x_n \rightarrow 0$ as $n \rightarrow \infty$, for all $x_0 \in \mathbb{R}$ and $\tau > 0$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that $\frac{1}{2}|x_1 - x_0|^2 + \tau f(x_1) \leq \tau f(x_0)$. Deduce that for all integers $N > M \geq 0$ $$\frac{1}{2}\sum_{M < n \leq N}|x_n - x_{n-1}|^2 \leq \tau\left(f(x_M) - f(x_N)\right)$$ Deduce that $|x_{n+1} - x_n| \rightarrow 0$ as $n \rightarrow \infty$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x \in \mathbb{R}$ and $\tilde{x} := p_f(x)$. Show that for all $v \in \mathbb{R}$ and $t \in \mathbb{R}$ $$\tau f(\tilde{x}) + \frac{1}{2}|\tilde{x} - x|^2 \leq \tau f(\tilde{x} + tv) + \frac{1}{2}|\tilde{x} + tv - x|^2$$ Let also $y \in \mathbb{R}$, and $\tilde{y} := p_f(y)$. Deduce that $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x, y \in \mathbb{R}$, $\tilde{x} := p_f(x)$, $\tilde{y} := p_f(y)$. Show that the right-hand side in inequality $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$ admits the asymptotic expansion $2tv(\tilde{x} - x + y - \tilde{y}) + o(t)$ as $t \rightarrow 0$.

Consider the family of all circles whose centers lie on the straight line $y = x$. If this family of circles is represented by the differential equation $P y ^ { \prime \prime } + Q y ^ { \prime } + 1 = 0$, where $P , Q$ are functions of $x , y$ and $y ^ { \prime }$ (here $y ^ { \prime } = \frac { d y } { d x } , y ^ { \prime \prime } = \frac { d ^ { 2 } y } { d x ^ { 2 } }$), then which of the following statements is (are) true?
(A) $P = y + x$
(B) $P = y - x$
(C) $P + Q = 1 - x + y + y ^ { \prime } + \left( y ^ { \prime } \right) ^ { 2 }$
(D) $P - Q = x + y - y ^ { \prime } - \left( y ^ { \prime } \right) ^ { 2 }$

The differential equation of all circles passing through the origin and having their centres on the $x$-axis is
(1) $x ^ { 2 } = y ^ { 2 } + x y \frac { d y } { d x }$
(2) $x ^ { 2 } = y ^ { 2 } + 3 x y \frac { d y } { d x }$
(3) $y ^ { 2 } = x ^ { 2 } + 2 x y \frac { d y } { d x }$
(4) $y ^ { 2 } = x ^ { 2 } - 2 x y \frac { d y } { d x }$

Statement-1: The slope of the tangent at any point P on a parabola, whose axis is the axis of $x$ and vertex is at the origin, is inversely proportional to the ordinate of the point P. Statement-2: The system of parabolas $y ^ { 2 } = 4 a x$ satisfies a differential equation of degree 1 and order 1.
(1) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
(2) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
(3) Statement-1 is true; Statement-2 is false.
(4) Statement-1 is false; Statement-2 is true.

If the differential equation representing the family of all circles touching $x$-axis at the origin is $\left( x ^ { 2 } - y ^ { 2 } \right) \frac { d y } { d x } = g ( x ) y$, then $g ( x )$ equals
(1) $\frac { 1 } { 2 } x ^ { 2 }$
(2) $2 x$
(3) $\frac { 1 } { 2 } x$
(4) $2 x ^ { 2 }$

A differential equation representing the family of parabolas with axis parallel to $y$-axis and whose length of latus rectum is the distance of the point $( 2 , - 3 )$ from the line $3 x + 4 y = 5$, is given by: (1) $11 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 10$ (2) $11 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 10$ (3) $10 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 11$ (4) $10 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 11$

The differential equation of the family of circles passing through the points $( 0,2 )$ and $( 0 , - 2 )$ is
(1) $2 x y \frac { d y } { d x } + \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$
(2) $2 x y \frac { d y } { d x } + \left( x ^ { 2 } + y ^ { 2 } - 4 \right) = 0$
(3) $2 x y \frac { d y } { d x } + \left( y ^ { 2 } - x ^ { 2 } + 4 \right) = 0$
(4) $2 x y \frac { d y } { d x } - \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$

The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :
(1) $\left( x ^ { 2 } - y ^ { 2 } + 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } - y ^ { 2 } - 2 x y \right) \mathrm { d } y$
(2) $\left( x ^ { 2 } + y ^ { 2 } + 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } + y ^ { 2 } - 2 x y \right) \mathrm { d } y$
(3) $\left( x ^ { 2 } + y ^ { 2 } - 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } + y ^ { 2 } + 2 x y \right) \mathrm { d } y$
(4) $\left( x ^ { 2 } - y ^ { 2 } + 2 x y \right) \mathrm { d } x = \left( x ^ { 2 } - y ^ { 2 } + 2 x y \right) \mathrm { d } y$

Let $f ( x )$ be a positive function such that the area bounded by $y = f ( x ) , y = 0$ from $x = 0$ to $x = a > 0$ is $e ^ { - a } + 4 a ^ { 2 } + a - 1$. Then the differential equation, whose general solution is $y = c _ { 1 } f ( x ) + c _ { 2 }$, where $c _ { 1 }$ and $c _ { 2 }$ are arbitrary constants, is
(1) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$
(2) $\left( 8 e ^ { x } - 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$
(3) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - \frac { d y } { d x } = 0$
(4) $\left( 8 e ^ { x } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } = 0$