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### Solved Examples

**Question 1:**Car A began a journey from a point at 9 am, traveling at 30 mph. At 10 am car B started traveling from the same point at 40 mph in the same direction as car A. At what time will car B pass car A?

**Solution:**

Let after t hours the distances D

=> D

Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it.

After (t - 1) hours, distance D

=> D

When car B passes car A, they are at the same distance from the starting point and therefore

=> 30 t = 40 (t - 1)

Solve the equation for t,

=> 30 t = 40t - 40

=> 10 t = 40

=> t = 4

=>

_{1}traveled by car A=> D

_{1}= 30 tCar B starts at 10 am and will therefore have spent one hour less than car A when it passes it.

After (t - 1) hours, distance D

_{2}traveled by car B=> D

_{2}= 40 (t - 1)When car B passes car A, they are at the same distance from the starting point and therefore

**D**_{1}= D_{2}=> 30 t = 40 (t - 1)

Solve the equation for t,

=> 30 t = 40t - 40

=> 10 t = 40

=> t = 4

=>

**Car B passes car A at = 9 + 4 = 13 pm.****Question 2:**Four times the first of three consecutive even integers is six more than the product of two and the third integer. Find the integers.

**Solution:**

First integer = x

Second integer = x + 2

Third integer = x + 4

Since four times the first integer equals six more than the product of two and the third integer.

=> 4x = 6 + 2(x + 4)

=> 4x = 6 + 2x + 8

=> 2x = 14

=> x = 7.

Hence,

First integer = x = 7

Second integer = x + 2 = 7 + 2 = 9

Third integer = x + 4 = 7 + 4 = 11.

Second integer = x + 2

Third integer = x + 4

Since four times the first integer equals six more than the product of two and the third integer.

=> 4x = 6 + 2(x + 4)

=> 4x = 6 + 2x + 8

=> 2x = 14

=> x = 7.

Hence,

First integer = x = 7

Second integer = x + 2 = 7 + 2 = 9

Third integer = x + 4 = 7 + 4 = 11.

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### Solved Examples

**Question 1:**Find the general solution for differential equation

(D

^{4}- 5D

^{3}+ 5D

^{2}+ 5D - 6)y = 0

**Solution:**

Given differential equation, (D

=> For general solution of equation,

Solve D

=> D

=> D

=> (D

Now

D

Factors of D

D

= D(D - 2) - 3(D - 2)

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D

=> D = 3, 2, 1, -1 or D = -1, 1,, 2, 3

=> General solution of differential equation is,

^{4 }- 5D^{3}+ 5D^{2}+ 5D - 6)y = 0=> For general solution of equation,

Solve D

^{4 }- 5D^{3}+ 5D^{2}+ 5D - 6 = 0=> D

^{4 }- 5D^{3}+ 6D^{2}- D^{2 }+ 5D - 6 = 0=> D

^{2}(D^{2}- 5D + 6) - (D^{2}- 5D + 6) = 0=> (D

^{2}- 5D + 6)(D^{2}- 1) = 0 ................................(1)Now

D

^{2 }- 1 = (D - 1)(D + 1) andFactors of D

^{2}- 5D + 6D

^{2}- 5D + 6 = D^{2}- 2D - 3D + 6= D(D - 2) - 3(D - 2)

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D

^{2}- 5D + 6)(D^{2}- 1) = (D - 3)(D - 2)(D - 1)(D + 1) = 0=> D = 3, 2, 1, -1 or D = -1, 1,, 2, 3

=> General solution of differential equation is,

=>

=>

**y = C**._{1}e^{-x}+ C_{2}e^{x}+ C_{3}e^{2x}+ C_{4}e^{3x}**Question 2:**Find the 20th term and the sum of the first 14 terms of this progression, 2, 5, 8, 11,.........?

**Solution:**

Given progression, 2, 5, 8, 11,......... is a arithmetic progression

Here First term (a) = 2

and Common difference (d) = 5 - 2 = 3

We know, n

and sum of n terms = $\frac{n}{2}${2a + (n - 1)d}

To find 20th term of the series

=> a

= 2 + 19 * 3

= 2 + 57

= 59

=>

Sum of first 14 terms = $\frac{14}{2}${2 * 2 + (14 - 1) * 3}

= 7{4 + 39}

= 7 * 43

= 301

=>

Here First term (a) = 2

and Common difference (d) = 5 - 2 = 3

We know, n

^{th}term can be represented as, a_{n}= a + (n - 1)dand sum of n terms = $\frac{n}{2}${2a + (n - 1)d}

To find 20th term of the series

=> a

_{20}= a + 19 d= 2 + 19 * 3

= 2 + 57

= 59

=>

**20th term = 59**Sum of first 14 terms = $\frac{14}{2}${2 * 2 + (14 - 1) * 3}

= 7{4 + 39}

= 7 * 43

= 301

=>

**Sum of first 14 terms = 301.**