(a) Show that the solution of the differential equation $(1+x^2)\dfrac{dy}{dx}=1+y^2$ is $\tan^{-1}y=\tan^{-1}x+C$ (or) $\tan^{-1}x=\tan^{-1}y+C$. **OR** (b) Prove $p\to(q\to r)\equiv(p\wedge q)\to r$ using truth table.

> [!TLDR] > (a) solves the given first-order differential equation by direct separation of variables; (b) proves a standard logical equivalence exhaustively with an 8-row truth table. > > **(a) Solving (1+x^2) dy/dx = 1+y^2** > 1. Given: $(1+x^2)\dfrac{dy}{dx}=1+y^2$. > 2. Separate the variables: $\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}$. > 3. Integrate both sides: $\displaystyle\int\dfrac{dy}{1+y^2}=\int\dfrac{dx}{1+x^2}$. > 4. $\tan^{-1}y=\tan^{-1}x+C$ (equivalently $\tan^{-1}x=\tan^{-1}y+C$, absorbing the sign of the constant), which is the required solution. > > **(b) Proving p -> (q -> r) is equivalent to (p AND q) -> r** > 1. Build the truth table over all 8 combinations of $p,q,r$: > > | $p$ | $q$ | $r$ | $q\to r$ | $p\to(q\to r)$ | $p\wedge q$ | $(p\wedge q)\to r$ | > |---|---|---|---|---|---|---| > | T | T | T | T | T | T | T | > | T | T | F | F | F | T | F | > | T | F | T | T | T | F | T | > | T | F | F | T | T | F | T | > | F | T | T | T | T | F | T | …

(a) Show that the solution of the differential equation (1+x²)(dy)/(dx)=1+y² is tan⁻¹y=tan⁻¹x+C (or) tan⁻¹x=ta

Question 114 of 126

(1+x^2)\dfrac{dy}{dx}=1+y^2

(b) Prove

p\to(q\to r)\equiv(p\wedge q)\to r

using truth table.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2022Subjective· 5mImportance★★★★★

90% · 114/126 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Variables Separable

A first-order differential equation is separable if it can be rearranged so that every $y$ (and $dy$ ) is on one side and every $x$ (and $dx$ ) is on the other — that is, written as $h(y)\,y'=g(x)$ , or equivalently as $f_1(x)g_1(y)\,dx+f_2(x)g_2(y)\,dy=0$ .

Method.

Rearrange the given equation into the separated form

\dfrac{f_1(x)}{f_2(x)}\,dx=-\dfrac{g_2(y)}{g_1(y)}\,dy.

Integrate both sides independently:

\int\dfrac{f_1(x)}{f_2(x)}\,dx=-\int\dfrac{g_2(y)}{g_1(y)}\,dy+C.

Only one arbitrary constant $C$ is needed — the two constants that would arise from integrating each side separately combine into a single overall constant.

3. If an initial condition is given (e.g. $y=y_0$ at $x=x_0$ ), substitute it into the integrated equation to evaluate $C$ and obtain the particular solution.

"Solving" a differential equation is therefore also called "integrating" it, since the whole process reduces to two ordinary integrations once the variables are separated.

Recognising a separable equation in disguise. Many equations that do not look separable at first become separable after a short algebraic step:

Product-to-sum trig identities — e.g. $\cos(x+y)+\cos(x-y)=2\cos x\cos y$ — turn a mixed trigonometric equation into one where $x$ and $y$ split apart.
Recognising an exact differential — noticing that $\dfrac{y\,dx-x\,dy}{y^2}=d\!\left(\dfrac{x}{y}\right)$ lets an equation be separated in the single combined variable $\dfrac{x}{y}$ directly, without a full substitution. …

Start your 14-day free trial to unlock the full solution →

(a) solves the given first-order differential equation by direct separation of variables; (b) proves a standard logical equivalence exhaustively with an 8-row truth table.

(a) Solving (1+x^2) dy/dx = 1+y^2

Given: $(1+x^2)\dfrac{dy}{dx}=1+y^2$ .
Separate the variables: $\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}$ .
Integrate both sides: $\displaystyle\int\dfrac{dy}{1+y^2}=\int\dfrac{dx}{1+x^2}$ .
$\tan^{-1}y=\tan^{-1}x+C$ (equivalently $\tan^{-1}x=\tan^{-1}y+C$ , absorbing the sign of the constant), which is the required solution.

(b) Proving p -> (q -> r) is equivalent to (p AND q) -> r

Build the truth table over all 8 combinations of $p,q,r$ :

$p$	$q$	$r$	$q\to r$	$p\to(q\to r)$	$p\wedge q$	$(p\wedge q)\to r$
T	T	T	T	T	T	T
T	T	F	F	F	T	F
T	F	T	T	T	F	T
T	F	F	T	T	F	T
F	T	T	T	T	F	T

Start your 14-day free trial to unlock the full solution →

←Previous questionQ32 — Solve : (dy)/(dx)=dfrac√(1-y²)√(1-x²)→Next questionQ34 — The general solution of the differential equation (dy)/(dx)=(y)/(x) is…

Q.(a) Show that the solution of the differential equation (1+x2)dydx=1+y2(1+x^2)\dfrac{dy}{dx}=1+y^2(1+x2)dxdy​=1+y2 is tan⁡−1y=tan⁡−1x+C\tan^{-1}y=\tan^{-1}x+Ctan−1y=tan−1x+C (or) tan⁡−1x=tan⁡−1y+C\tan^{-1}x=\tan^{-1}y+Ctan−1x=tan−1y+C. OR

🎓Unlock everything free for 14 days

Q.(a) Show that the solution of the differential equation $(1+x^2)\dfrac{dy}{dx}=1+y^2$ is $\tan^{-1}y=\tan^{-1}x+C$ (or) $\tan^{-1}x=\tan^{-1}y+C$ . OR

Unlock everything free for 14 days