verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation y = ex + 1 : y″ – y′ = 0 - general and particular solutions of a differential equation
verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation y = ex + 1 : y′ – y′ = 0
y = ex + 1 differentiating both sides of this equation with respect to x, we get: thus, the given function is the solution of the corresponding differential equation.
i'm going to assume that "directly related" means a linear relationship. otherwise, there is no way to write an equation based on the information provided.assuming a linear function, use the form y = mx + b. with the information given, m and b cannot be determined. however, the question is simply the domain. for a linear equation, the domain is simply "all real numbers".in the space provided, type the answer without any spaces between the letters, numbers, or symbols.i have no idea how that relates to the stated question....... let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) general systems thursday, september 29, 2016 3:50 pm a fruit store sell two types of fruit, apples and oranges. apples are $2 and oranges are $3.today 50 fruit was sold and the store made $110. 110 = 2▯ + 3▯ 50 = ▯ + ▯ ▯ = 40 ▯ = 10 general ▯▯ ▯+ ▯ ▯▯ ▯⋯+ ▯ ▯ = ▯ ▯ a linear equation is an equation in the form of: • x's are variables and a's are constants. • solution set for a linear equation is the set of all solutions. • set of linear equations where a commonsolution is desired is called a system of linear equations. graphs • 1 variables is a point • 2 variables is a line • 3 variables is a plane • 4+ variables is a hyperplane examples of systems 2 equations, 2 unknowns • no solutions • one hint- here, we will be proceeding by differentiating the given function with respect to x and then putting $\dfrac{{dy}}{{dx}} = 0$ and afterwards substituting the given extreme values of x to obtain the two equation in terms of m and n only and then we will solve them.given, $y = m\log x + n{x^2} + x{\text{ }} \to {\text{(1)}}$ which is a function in terms of variable x.it is also given that the function represented in equation (1) has its extreme values at $x = 2$ and $x = 1$.as we know that extreme values of any function y are obtained by finding $\dfrac{{dy}}{{dx}}$ and then putting $\dfrac{{dy}}{{dx}} = 0$.by differentiating the function given by equation (1) both sides with respect to x, we get$\dfrac{{dy}}{{dx}} = \dfrac{{d\left[ {m\log x + n{x^2} + x} \right]}}{{dx}} = \dfrac{{d\left( {m\log x} \right)}}{{dx}} + \dfrac{{d\left( {n{x^2}} \right)}}{{dx}} + \dfrac{{dx}}{{dx}}$since, m and n are constants (i.e., independent of variable x) so we can take them out of the differentiation.\[   \rightarrow \dfrac{{dy}}{{dx}} = m\left[ {\dfrac{{d\left( {\log x} \right)}}{{dx}}} \right] + n\left[ {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right] + 1 = m\left[ {\dfrac{1}{x}} \right] + n\left[ {2x} \right] + 1 \\   \rightarrow \dfrac{{dy}}{{dx}} = \dfrac{m}{x} + 2nx + 1 \\  \]put $\dfrac{{dy}}{{dx}} = 0$ for the extreme values of the variable x, we get\[   \rightarrow 0 = \dfrac{m}{x} + 2nx + 1 \\   \rightarrow 0 = \dfrac{{m + 2n{x^2} + x}}{x} \\   \rightarrow 0 = m + 2n{x^2} + x \\   \rightarrow 2n{x^2} + x + m = 0{\text{ }} \to {\text{(2)}} \\  \]so, we are having a quadratic equation in variable x which means there will be two extreme values of x. these extreme values are $x = 2$ and $x = 1$. these two given extreme values should satisfy the quadratic equation given by equation (2).put x=2 in equation (2), we get\[   \rightarrow 2n{\left( 2 \right)^2} + 2 + m = 0 \\   \rightarrow 8n + 2 + m = 0{\text{ }} \to {\text{(3)}} \\  \]put x=1 in equation (2), we get\[   \rightarrow 2n{\left( 1 \right)^2} + 1 + m = 0 \\   \rightarrow 2n + 1 + m = 0{\text{ }} \to {\text{(4)}} \\  \]in order to obtain the values of constants m and n we will subtract equation (4) from equation (3), we have\[   \rightarrow 8n + 2 + m - \left( {2n + 1 + m} \right) = 0 - 0 \\   \rightarrow 8n + 2 + m - 2n - 1 - m = 0 \\   \rightarrow 6n + 1 = 0 \\   \rightarrow 6n = - 1 \\   \rightarrow n = \dfrac{{ - 1}}{6} \\  \]put \[n = \dfrac{{ - 1}}{6}\] in equation (4) to obtain the value of m, we have\[   \rightarrow 2\left[ {\dfrac{{ - 1}}{6}} \right] + 1 + m = 0 \\   \rightarrow \dfrac{{ - 1}}{3} + 1 + m = 0 \\   \rightarrow m = \dfrac{1}{3} - 1 = \dfrac{{1 - 3}}{3} \\   \rightarrow m = \dfrac{{ - 2}}{3} \\  \]put \[m = \dfrac{{ - 2}}{3}\] and \[n = \dfrac{{ - 1}}{6}\] to find the value for the expression $2m + 10n$, we get$  2m + 10n = 2\left[ {\dfrac{{ - 2}}{3}} \right] + 10\left[ {\dfrac{{ - 1}}{6}} \right] = \dfrac{{ - 4}}{3} - \dfrac{5}{3} = \dfrac{{ - 4 - 5}}{3} = \dfrac{{ - 9}}{3} \\   \rightarrow 2m + 10n = - 3 \\  $hence, option e is correct.note- at the extreme values for any function y, $\dfrac{{dy}}{{dx}} = 0$. also, if $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ that means at this extreme value, minima occurs (i.e., we will be getting the minimum value of y corresponding to this extreme value of x) and if $\dfrac{{{d^2}y}}{{d{x^2}}} if statements always check if an expression evaluates to the value of true. since you’re checking whether the value true  value true then it’ll run every time. as for the “purpose” of the if statement, it’s here to create a “block scope” within the “function scope”.

interestingly, if you remove … equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1. given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6) Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4].