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APPLICATIONS OF INDEFINITE INTEGRALS
Midterm Project in Calculus 2
2
nd
Sem., A.Y. 2019-2020
Submitted by:
Paul Allen Jarilla
CE-1B
Submitted to:
Roderick B. Astillero
Associate Professor II
Showing Page:
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1. Find the particular solution of the differential equation,
  satisfying
the initial conditions that when 
SOLUTION
GRAPH
Simplify expression;

󰆒

󰆒


󰆒

󰆒


󰆒

󰆒


󰆒
󰇛

󰇜


󰇛
󰇜
󰇛
󰇜

󰇛
󰇜
󰇛
󰇜
󰆒

󰇛

󰇜
; or



 

 

Integrate both sides;


 


 
General solution
Evaluate: x=4, y=9
󰇛
󰇜
󰇛
󰇜

󰇛
󰇜


󰇛
󰇜

󰇛
󰇜



󰇛
󰇜


󰇛
󰇜


 


󰇛
󰇜
Particular solution
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2. Find the particular solution of the differential equation,
󰆒

satisfying the initial
conditions that when 
SOLUTION
GRAPH
󰆒





󰇛

󰇜

Integrate both sides;



 


General Solution
Evaluate; x=3, y=35

󰇛
󰇜
󰇛
󰇜
  


Particular Solution
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3. Find the particular solution of the differential equation,

󰆓

󰇛

󰇜

satisfying
the initial conditions that when 
SOLUTION
GRAPH
Simplify:

󰆒

󰇛

󰇜


󰆒

󰇛

󰇜


󰇧
󰇛
󰆒
󰇜
󰇨
󰇛

󰇜
󰆒





󰇛

󰇜

Integrate both sides;
 
 


General Solution
Evaluate: x=2; y=20

󰇛
󰇜
󰇛
󰇜
 



Particular Solution
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4. Find the particular solution of the differential equation,

󰆓

󰆓

󰇛

󰇜
satisfying the initial conditions that when

󰇡
󰇢

SOLUTION
Simplify:

󰆒
󰆒

󰇛

󰇜

󰆒
󰆒
󰇛

󰇜
󰆒
󰇛

󰇜

󰇛

󰇜
󰆒

󰇛

󰇜
󰇛

󰇜
Integrate both sides:

󰇛

󰇜
󰇛

󰇜

󰇛

󰇜







󰇛

󰇜

General Solution
Evaluate: x=

󰇡
󰇢
, y=12


󰇡
󰇢





󰇛

󰇜

Particular Solution
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5. Find the particular solution of the differential equation,

󰆒
 satisfying
the initial conditions that when
󰇡
󰇢
SOLUTION
GRAPH
Simplify:

󰆒


󰆒


󰆒

󰆒
󰆒
Integrate both sides:







󰇛
󰇜
General Solution
Evaluate:
󰇡
󰇢
󰇡

󰇛

󰇜
󰇢


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APPLICATIONS OF INDEFINITE INTEGRALS Midterm Project in Calculus 2 2nd Sem., A.Y. 2019-2020 Submitted by: Paul Allen Jarilla CE-1B Submitted to: Roderick B. Astillero Associate Professor II 1. Find the particular solution of the differential equation, 𝑥 2 + 4𝑥𝑦 − 20𝑦 = 20 satisfying the initial conditions that when 𝑥 = 4 , 𝑦 = 9. SOLUTION Simplify expression; 𝑥 2 + 4𝑥𝑦 ′ − 20𝑦 ′ = 20 𝑥 2 + 4𝑥𝑦 ′ − 20𝑦 ′ − 𝑥 2 = 20 − 𝑥 2 4𝑥𝑦 ′ − 20𝑦 ′ = 20 − 𝑥 2 4𝑦 ′(𝑥−5) = 20 − 𝑥 2 4𝑦′(𝑥 − 5) 20 𝑥2 = − 4(𝑥 − 5) 4(𝑥 − 5) 4(𝑥 − 5) 20−𝑥 2 𝑥 5 5 𝑦 ′ = 4(−5+𝑥) ; or 𝑦 = − 4 − 4 − 4𝑥−20 𝑑𝑦 𝑥 5 5 =− − − 𝑑𝑥 4 4 4𝑥 − 20 𝑥 5 5 ) 𝑑𝑥 𝑑𝑦 = (− − − 4 4 4𝑥 − 20 Integrate both sides; 𝑥 5 5 𝑦 = − ∫ 𝑑𝑥 − ∫ 𝑑𝑥 − ∫ 𝑑𝑥 4 4 4𝑥 − 20 𝑥2 5 5 𝑦 = − − 𝑥 − ln|4𝑥 − 20| + 𝐶 8 4 4 General solution Evaluate: x=4, y=9 (4)2 5 5 9=− − (4) − ln|4(4) − 20| + 𝑐 8 4 4 5 9 = −(2) − 5 − (𝑙𝑛(4)) + 𝑐 4 5 −𝑐 = −16 − 𝑙𝑛(4) 4 5 𝑐 = 16 + 𝑙𝑛(4) ≈ 17.73 4 𝑥2 5 5 5 𝑦 = − − 𝑥 − ln|4𝑥 − 20| + [16 + 𝑙𝑛(4)] 8 4 4 4 Particular solution GRAPH 2. Find the particular solution of the differential equation, 𝑦 ′ = 2𝑥 2 + 5 satisfying the initial conditions that when 𝑥 = 3 , 𝑦 = 35 . SOLUTION 𝑦 ′ = 2𝑥 2 + 5 𝑑𝑦 = 2𝑥 2 + 5 𝑑𝑥 𝑑𝑦 = (2𝑥 2 + 5)𝑑𝑥 Integrate both sides; 𝑦 = ∫ 2𝑥 2 + 5𝑑𝑥 = ∫ 2𝑥 2 𝑑𝑥 + ∫ 5𝑑𝑥 2𝑥 3 + 5𝑥 + 𝐶 3 General Solution 𝑦= Evaluate; x=3, y=35 2(3)3 35 = + 5(3) + 𝐶 3 35 − 18 − 15 = 𝐶 𝐶=2 2𝑥 3 + 5𝑥 + 2 3 Particular Solution 𝑦= GRAPH 3. Find the particular solution of the differential equation, the initial conditions that when 𝑥 = 2 , 𝑦 = 20. SOLUTION Simplify: 16𝑥 𝑒 𝑦 ′ − 48𝑥 (𝑒+1) = 48𝑥 𝑒 3 16𝑥 𝑒 𝑦 ′ − 16𝑥 (𝑒+1) = 48𝑥 𝑒 3 (𝑦 ′ ) 𝑒 16𝑥 ( − 𝑥) = 3(16𝑥 𝑒 ) 3 𝑦 ′ = 3𝑥 + 9 𝑑𝑦 = 3𝑥 + 9 𝑑𝑥 𝑑𝑦 = (3𝑥 + 9)𝑑𝑥 Integrate both sides; 𝑦 = ∫ 3𝑥 + 9𝑑𝑥 = ∫ 3𝑥𝑑𝑥 + ∫ 9𝑑𝑥 3𝑥 2 𝑦= + 9𝑥 + 𝐶 2 General Solution Evaluate: x=2; y=20 3(2)2 20 = + 9(2) + 𝐶 2 20 − 6 − 18 = 𝐶 𝐶 = −4 3𝑥 2 𝑦= + 9𝑥 − 4 2 Particular Solution 16𝑥 𝑒 𝑦 ′ −48𝑥 (𝑒+1) GRAPH 3 = 48𝑥 𝑒 satisfying 4. Find the particular solution of the differential equation, √ −1 satisfying the initial conditions that when 𝑥 = √ 3 + SOLUTION 3𝑥 2 𝑦 ′ + 𝑦 ′ = 𝑙𝑛(3𝑥 2 + 1) 𝑥 3𝑥 2 𝑦 ′ + 𝑦 ′ = 𝑙𝑛2 (3𝑥 2 + 1) 𝑥 𝑦 ′ (3𝑥 2 + 1) = 𝑥𝑙𝑛2 (3𝑥 2 + 1) 𝑥𝑙𝑛2 (3𝑥 2 + 1) ′ 𝑦 = (3𝑥 2 + 1) √ Integrate both sides: 𝑥𝑙𝑛2 (3𝑥 2 + 1) 𝑦=∫ 𝐿𝑒𝑡𝑢 = 𝑙𝑛(3𝑥 2 + 1) (3𝑥 2 + 1) 6𝑥𝑑𝑥 𝑑𝑢 𝑥𝑑𝑥 𝑑𝑢 = 2 ; = 2 3𝑥 + 1 6 3𝑥 + 1 𝑑𝑢 2 =∫ 𝑢 6 1 𝑢3 = ⋅ +𝐶 6 3 𝑙𝑛3 (3𝑥 2 + 1) = +𝐶 18 General Solution 3 (3 3√6) 3 ≈ 8.795, y=12 2 3 (3 √6) −1 𝑒 𝑙𝑛3 3 (√ + 3 3 ( 12 = 18 162 12 − =𝐶 18 12 − 9 = 𝐶 𝐶=3 𝑙𝑛3 (3𝑥 2 + 1) 𝑦= +3 18 Particular Solution 3 (3 √6) 3 , 𝑦 = 12 . GRAPH Simplify: −1 𝑒 Evaluate: x=√ + 𝑒 3𝑥 2 𝑦 ′ +𝑦 ′ ) +1 )+𝐶 𝑥 = 𝑙𝑛(3𝑥 2 + 1) 5. Find the particular solution of the differential equation, √3𝑦 ′ √1 − 𝑥 4 − √3𝑥 = 0 satisfying 𝜋 the initial conditions that when 𝑥 = √𝑠𝑖𝑛 ( 9 ) ≈ 0.585, 𝑦 = 7. SOLUTION Simplify: √3𝑦 ′ √1 − 𝑥 4 − √3𝑥 = 0 √3𝑦 ′ √1 − 𝑥 4 = √3𝑥 3𝑦 ′ √1 − 𝑥 4 = 3x 𝑦 ′ √1 − 𝑥 4 = x x 𝑦′ = √1 − 𝑥 4 Integrate both sides: x y=∫ √1 − 𝑥 4 𝑑𝑢 𝐿𝑒𝑡 𝑎 = 1; 𝑢 = 𝑥 2 ; 𝑑𝑢 = 2𝑥𝑑𝑥; = 𝑥𝑑𝑥 2 𝑑𝑢 2 =∫ √𝑎2 − 𝑢2 1 = 𝐴𝑟𝑐𝑠𝑖𝑛(𝑥 2 ) + 𝐶 2 General Solution 𝜋 Evaluate: 𝑥 = √𝑠𝑖𝑛 ( ) ≈ 0.585, 𝑦 = 7 9 2 1 7 = 𝐴𝑟𝑐𝑠𝑖𝑛 ((√𝑠𝑖𝑛(10)) ) + 𝐶 2 7 − 10 = 𝐶 𝐶 = −3 GRAPH Name: Description: ...
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