Consider the differential equation $\frac { d y } { d x } = \frac { 3 x ^ { 2 } } { e ^ { 2 y } }$.
(a) Find a solution $y = f ( x )$ to the differential equation satisfying $f ( 0 ) = \frac { 1 } { 2 }$.
(b) Find the domain and range of the function $f$ found in part (a).

Which of the following is the solution to the differential equation $\frac { d y } { d x } = 2 \sin x$ with the initial condition $y ( \pi ) = 1$ ?
(A) $y = 2 \cos x + 3$
(B) $y = 2 \cos x - 1$
(C) $y = - 2 \cos x + 3$
(D) $y = - 2 \cos x + 1$
(E) $y = - 2 \cos x - 1$

A curve $C$ has the property that the slope of the tangent at any given point $( x , y )$ on $C$ is $\frac { x ^ { 2 } + y ^ { 2 } } { 2 x y }$. a) Find the general equation for such a curve. Possible hint: let $z = \frac { y } { x }$. b) Specify all possible shapes of the curves in this family. (For example, does the family include an ellipse?)

We are given $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ and $g : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ two functions continuous and 1-periodic in each of their arguments, and $\omega \in \mathbb { R } ^ { 2 } , x \in \mathbb { R }$ two parameters. We consider the following problem $$\left\{ \begin{array} { l } F ^ { \prime } ( t ) = f ( \alpha ( t ) ) \\ \alpha ^ { \prime } ( t ) = \omega + x g ( \alpha ( t ) ) \end{array} \right.$$ with the initial conditions $F ( 0 ) = 0$ and $\alpha ( 0 ) = ( 0,0 )$, where $\alpha : \mathbb { R } \rightarrow \mathbb { R } ^ { 2 }$ and $F : \mathbb { R } \rightarrow \mathbb { C }$ are the unknown functions. We assume that $f$ has zero average, that is $\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } f \left( \theta _ { 1 } , \theta _ { 2 } \right) d \theta _ { 1 } d \theta _ { 2 } = 0$.
We assume $x = 0$. Determine the unique solution $( F , \alpha )$ of system (3) with initial conditions $F ( 0 ) = 0$ and $\alpha ( 0 ) = ( 0,0 )$.

We are given $f : \mathbb { R } ^ { 2 } \rightarrow \mathbb { C }$ and $g : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ two functions continuous and 1-periodic in each of their arguments, and $\omega \in \mathbb { R } ^ { 2 }$. We consider the system $$\left\{ \begin{array} { l } F ^ { \prime } ( t ) = f ( \alpha ( t ) ) \\ \alpha ^ { \prime } ( t ) = \omega \end{array} \right.$$ with $F(0)=0$, $\alpha(0)=(0,0)$. We assume that $f$ has zero average. The vector $\omega = \left( \omega _ { 1 } , \omega _ { 2 } \right)$ is said to be resonant if there exists $\left( k _ { 1 } , k _ { 2 } \right) \in \mathbb { Z } ^ { 2 } \backslash \{ ( 0,0 ) \}$ such that $k _ { 1 } \omega _ { 1 } + k _ { 2 } \omega _ { 2 } = 0$.
Show that, if $\omega$ is resonant, there exists a function $f$ for which $F ( t ) = t$.

We consider the Cauchy problem associated with $F_0$ defined by: $$\forall y \in ]0, +\infty[, \quad F_0(y) = ay\ln\left(\frac{\theta}{y}\right)$$ with $a, \theta > 0$ and $0 < y_{\text{init}} < \theta$. We denote $\phi_0$ the solution of this problem on $[0, +\infty[$.
(a) Show that there exists $\varepsilon > 0$ such that for all $t \in ]0, \varepsilon]$ we have $y_{\text{init}} < \phi_0(t) < \theta$.
(b) By considering the function $z_0(t) = \ln\left(\phi_0(t)/\theta\right)$ find the expression of $\phi_0$.
(c) Deduce that $\phi_0$ satisfies $y_{\text{init}} < \phi_0(t) < \theta$ for all $t \in ]0, +\infty[$ and that moreover $\phi_0$ is strictly increasing.

For $0 < \mu \leqslant 1$, we consider $F_\mu$ defined by: $$\forall y \in ]0, +\infty[, \quad F_\mu(y) = \frac{a}{\mu} y\left(1 - \left(\frac{y}{\theta}\right)^\mu\right)$$ with $a, \theta > 0$ and $0 < y_{\text{init}} < \theta$. By considering the function $z_\mu(t) = \phi_\mu(t)^{-\mu}$ find the expression of the solution $\phi_\mu$ on $[0, +\infty[$ associated with $F_\mu$.

Suppose that $S _ { 0 } = 0$. Give the expression of the solution triplet $( S , I , R )$ of system $( F )$.
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$ with $S_0 + I_0 + R_0 = 1$ and $S_0, I_0, R_0 \in [0,1]$.

With $S _ { 0 } = 1 / 2$, $I _ { 0 } = 1 / 2$, $R _ { 0 } = 0$, and $h : \mathbf { R } _ { + } \rightarrow \mathbf { R }$ defined by
$$\forall x \in \mathbf { R } _ { + } \quad h ( x ) = \ln \left( \frac { S ( x ) } { S _ { 0 } } \right) = \ln ( 2 S ( x ) ),$$
show that $h$ is a solution of the Cauchy problem $(C)$:
$$( C ) : \left\{ \begin{array} { l } y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) } \\ y ( 0 ) = 0 \end{array} \right.$$
using the relation established in question 18.

In this question only, we set $f(x) := \frac{1}{2}Lx^2$ for all $x \in \mathbb{R}$, where $L > 0$ is fixed. a) Show that $x_{n+1} = (1 - \tau L)x_n$, then express directly $x_n$ as a function of $x_0$ and $n$. b) We suppose $x_0 \neq 0$. Justify that $x_n \rightarrow 0$ if and only if $0 < \tau < 2/L$.

11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.
Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$ or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$ Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$ $\frac { d Y } { d X } = X + \frac { Y } { X }$ $\frac { d Y } { d X } - \frac { Y } { X } = X$ [Figure] I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$ $\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$. It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$. So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$ $\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$. ⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.

11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.
Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$ or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$ Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$ $\frac { d Y } { d X } = X + \frac { Y } { X }$ $\frac { d Y } { d X } - \frac { Y } { X } = X$ [Figure] I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$ $\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$. It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$. So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$ $\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$. ⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.

12. If length of tangent at any point on the curve $y = f ( x )$ intercepted between the point and the $x$-axis is of length 1 . Find the equation of the curve.

13. If $x d y = y ( d x + y d y ) , y ( 1 ) = 1$ and $y ( x ) > 0$. Then $y ( - 3 ) = :$
(a) 3
(b) 2
(c) 1
(d) 0

15. A tangent drawn to the curve $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ at $\mathrm { P } ( \mathrm { x } , \mathrm { y } )$ cuts the x -axis and y -axis at A and B respectively such that $\mathrm { BP } : \mathrm { AP } = 3 : 1$, given that $\mathrm { f } ( 1 ) = 1$, then
(A) equation of curve is $x \frac { d y } { d x } - 3 y = 0$
(B) normal at (1,1) is $x + 3 y = 4$
(C) curve passes through ( 2 , 1/8)
(D) equation of curve is $x \frac { d y } { d x } + 3 y = 0$

Sol. (C), (D)

Equation of the tangent is
$$\mathrm { Y } - \mathrm { y } = \frac { \mathrm { dy } } { \mathrm { dx } } ( \mathrm { X } - \mathrm { x } )$$
Given $\frac { \mathrm { BP } } { \mathrm { AP } } = \frac { 3 } { 1 }$ so that
$$\begin{aligned} & \Rightarrow \quad \frac { d x } { x } = - \frac { d y } { 3 y } \Rightarrow x \frac { d y } { d x } + 3 y = 0 \\ & \Rightarrow \quad \ln x = - \frac { 1 } { 3 } \ln y - \ln c \Rightarrow \ln x ^ { 3 } = - ( \ln c y ) \\ & \Rightarrow \quad \frac { 1 } { x ^ { 3 } } = c y . \text { Given } f ( 1 ) = 1 \Rightarrow c = 1 \\ & \therefore y = \frac { 1 } { x ^ { 3 } } . \end{aligned}$$
[Figure]

49. The differential equation $\frac { d y } { d x } = \frac { \sqrt { 1 - y ^ { 2 } } } { y }$ determines a family of circles with
(A) variable radii and a fixed centre at $( 0,1 )$
(B) variable radii and a fixed centre at $( 0 , - 1 )$
(C) fixed radius 1 and variable centres along the $x$-axis
(D) fixed radius 1 and variable centres along the $y$-axis Answer O O O O
(A)
(B)
(C)
(D)

Let a solution $y = y ( x )$ of the differential equation
$$x \sqrt { x ^ { 2 } - 1 } d y - y \sqrt { y ^ { 2 } - 1 } d x = 0$$
satisfy $y ( 2 ) = \frac { 2 } { \sqrt { 3 } }$. STATEMENT-1 : $y ( x ) = \sec \left( \sec ^ { - 1 } x - \frac { \pi } { 6 } \right)$ and STATEMENT-2 : $y ( x )$ is given by
$$\frac { 1 } { y } = \frac { 2 \sqrt { 3 } } { x } - \sqrt { 1 - \frac { 1 } { x ^ { 2 } } }$$
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True

Match the statements/expressions given in Column I with the values given in Column II.
Column I
(A) The number of solutions of the equation $$xe^{\sin x}-\cos x=0$$ in the interval $\left(0,\frac{\pi}{2}\right)$
(B) Value(s) of $k$ for which the planes $kx+4y+z=0,4x+ky+2z=0$ and $2x+2y+z=0$ intersect in a straight line
(C) Value(s) of $k$ for which $$|x-1|+|x-2|+|x+1|+|x+2|=4k$$ has integer solution(s)
(D) If $$y^{\prime}=y+1\text{ and }y(0)=1$$ then value(s) of $y(\ln2)$
Column II
(p) 1
(q) 2
(r) 3
(s) 4
(t) 5

Let f be a real-valued differentiable function on $\mathbf { R }$ (the set of all real numbers) such that $f ( 1 ) = 1$. If the $y$-intercept of the tangent at any point $P ( x , y )$ on the curve $y = f ( x )$ is equal to the cube of the abscissa of $P$, then the value of $f ( - 3 )$ is equal to

A curve passes through the point $\left( 1 , \frac { \pi } { 6 } \right)$. Let the slope of the curve at each point $( x , y )$ be $\frac { y } { x } + \sec \left( \frac { y } { x } \right) , x > 0$. Then the equation of the curve is
(A) $\quad \sin \left( \frac { y } { x } \right) = \log x + \frac { 1 } { 2 }$
(B) $\quad \operatorname { cosec } \left( \frac { y } { x } \right) = \log x + 2$
(C) $\quad \sec \left( \frac { 2 y } { x } \right) = \log x + 2$
(D) $\quad \cos \left( \frac { 2 y } { x } \right) = \log x + \frac { 1 } { 2 }$

A solution curve of the differential equation $\left(x^2 + xy + 4x + 2y + 4\right)\frac{dy}{dx} - y^2 = 0, x > 0$, passes through the point $(1,3)$. Then the solution curve
(A) intersects $y = x + 2$ exactly at one point
(B) intersects $y = x + 2$ exactly at two points
(C) intersects $y = (x+2)^2$
(D) does NOT intersect $y = (x+3)^2$

If $y = y ( x )$ satisfies the differential equation
$$8 \sqrt { x } ( \sqrt { 9 + \sqrt { x } } ) d y = ( \sqrt { 4 + \sqrt { 9 + \sqrt { x } } } ) ^ { - 1 } d x , \quad x > 0$$
and $y ( 0 ) = \sqrt { 7 }$, then $y ( 256 ) =$
[A] 3
[B] 9
[C] 16
[D] 80

Let $f : ( 0 , \pi ) \rightarrow \mathbb { R }$ be a twice differentiable function such that
$$\lim _ { t \rightarrow x } \frac { f ( x ) \sin t - f ( t ) \sin x } { t - x } = \sin ^ { 2 } x \text { for all } x \in ( 0 , \pi )$$
If $f \left( \frac { \pi } { 6 } \right) = - \frac { \pi } { 12 }$, then which of the following statement(s) is (are) TRUE?
(A) $f \left( \frac { \pi } { 4 } \right) = \frac { \pi } { 4 \sqrt { 2 } }$
(B) $f ( x ) < \frac { x ^ { 4 } } { 6 } - x ^ { 2 }$ for all $x \in ( 0 , \pi )$
(C) There exists $\alpha \in ( 0 , \pi )$ such that $f ^ { \prime } ( \alpha ) = 0$
(D) $f ^ { \prime \prime } \left( \frac { \pi } { 2 } \right) + f \left( \frac { \pi } { 2 } \right) = 0$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a differentiable function with $f ( 0 ) = 0$. If $y = f ( x )$ satisfies the differential equation
$$\frac { d y } { d x } = ( 2 + 5 y ) ( 5 y - 2 )$$
then the value of $\lim _ { x \rightarrow - \infty } f ( x )$ is $\_\_\_\_$ .

If $y ( x )$ is the solution of the differential equation
$$x d y - \left( y ^ { 2 } - 4 y \right) d x = 0 \text { for } x > 0 , \quad y ( 1 ) = 2$$
and the slope of the curve $y = y ( x )$ is never zero, then the value of $10 y ( \sqrt { 2 } )$ is $\_\_\_\_$ .