Definition: An equation of the form dy/dx + Py = Q, where P and Q are function of x only, is called a Linear differential equation.
Solution of Linear Equation: Given equation is dy/dx + P y = Q
\({{e}^{\int{p.dx}}}\left( \frac{dy}{dx}+Py \right)=Q.{{e}^{\int{p.dx}}}\),
\(\frac{d}{dx}y.{{e}^{\int{p.dx}}}=Q.{{e}^{\int{p.dx}}}\),
\(\frac{d}{dx}y.{{e}^{\int{p.dx}}}.dx=Q.{{e}^{\int{p.dx}}}.dx\),
\(y.{{e}^{\int{p.dx}}}=\int{Q.{{e}^{\int{p.dx}}}}dx+C\).
Example 1: Solve (1 + x²) dy/dx + y = e^{tan¯¹x}
Solution: Given that (1 + x²) dy/dx + y = e^{tan¯¹x}
\(\frac{dy}{dx}+\frac{1}{1+{{x}^{2}}}.y=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}\),
\(P=\frac{1}{1+{{x}^{2}}}\), \(Q=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}\),
Integration factor (IF) = \({{e}^{\int{\frac{1}{1+{{x}^{2}}}dx}}}={{e}^{{{\tan }^{-1}}x}}\),
Solution is \(y{{e}^{{{\tan }^{-1}}x}}=\int{\frac{{{e}^{2{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}\),
Put tan¯¹x = t
Then 1/ (1 + x²) dx = dt
\(y{{e}^{{{\tan }^{-1}}x}}=\frac{1}{2}{{e}^{2{{\tan }^{-1}}x}}+c\),
Example 2: Solve x (x – 1) dy/dx – y = x³ (x – 1)³
Solution: Given that x (x – 1) dy/dx – y = x³ (x – 1)³
\(\frac{dy}{dx}-\frac{1}{x(x-1)}y={{x}^{2}}{{(x-1)}^{2}}\),
P = -1/ x (x – 1), Q = x² (x – 1)²
\(\int{\frac{-dx}{x(x-1)}}=\int{\left( \frac{1}{x}-\frac{1}{x-1} \right)dx}\),
= log x – log (x – 1)
= log x/ (x – 1)
Integration factor (I.F) \(={{e}^{\int{-\frac{dx}{x(x-1)}}}}={{e}^{\log \frac{x}{x-1}}}\),
= x/ (x – 1)
Solution is y.x/ (x – 1) = ∫x³ (x – 1) dx = ∫ (x⁴ – x³) dx
= x⁵/5 – x⁴/4 + c
The solution is xy/ x – 1 = x⁵/5 – x⁴/4 + c.
]]>When a particle is thrown vertically upwards, it reaches a certain height and comes back to its original position. But when it is given greater initial velocity, it reaches greater height before coming back. A particle can be thrown up with a certain minimum initial velocity so that, the particle goes beyond the earth’s gravitational field and escape from earth, this velocity is known as escape velocity.
What is Escape Velocity?
Escape Velocity is the minimum velocity required by a body to be projected overcomes the gravitational pull of the earth. It is the minimum velocity required by an object to escape the gravitational field that is, escape the land without ever falling back. An object that has this velocity at the earth’s surface will totally escape the earth’s gravitational field ignoring the losses due to the atmosphere.
Escape Velocity Formula:
\(Escape\,\,Velocity({{V}_{e}})=\sqrt{\frac{2GM}{R}}\).
Where,
G = Universal Gravitational Constant = 6.673 x 10⁻¹¹ N.m²/ kg²
M = Mass of the planet (or) moon,
R = Radius from center of gravity.
How to find the Escape Velocity?
Problem: Find the escape velocity of the moon, if mass is 7.35 x 10²² kg and radius is 1.2 x 10⁵ m?
Solution: Given,
Mass (m) = 7.35 x 10²² kg,
Radius (R) = 1.2 x 10⁵ m,
G = 6.673 x 10⁻¹¹ N.m²/ kg².
We know that:
\(Escape\,\,Velocity({{V}_{e}})=\sqrt{\frac{2GM}{R}}\).
\(=\sqrt{\frac{2\times 6.673\times {{10}^{-11}}\times 7.35\times {{10}^{22}}}{1.2\times {{10}^{5}}}}\).
∴ Escape Velocity (V_{e}) = 9.04 x 10³ m/ sec.
]]>The time represented by the clock hands of a pendulum clock depends on the number of oscillations performed by pendulum. Every time it reaches to its extreme position. The second hand of the clock advances by one second that means second hand moves by two seconds when one oscillation is complete.
A pendulum clock keeps proper time at temperature θ₀. If temperature is increased to θ (> θ₀) then due to linear expansion, length of pendulum and hence its time period will increase. Now,
Let, \(T=2\pi \sqrt{\frac{{{L}_{0}}}{g}}\) at temperature θ₀ and \(T’=2\pi \sqrt{\frac{L}{g}}\) at temperature θ.
\(\frac{T’}{T}=\sqrt{\frac{L’}{L}}=\sqrt{\frac{L(1+\alpha \Delta \theta )}{L}}=1+\frac{1}{2}\alpha \Delta \theta \).
Therefore, change in time per unit time lapsed is: \(\frac{T’-T}{T}=\frac{1}{2}\alpha \Delta \theta \).
Gain or loss in time in duration of T is: ΔT = ½ αΔθT. If T is the correct time then,
θ < θ₀, T’ < T – clock becomes fast and gain time.
θ > θ₀, T’ > T – clock becomes slow and loses time.
Final change in time period, \(\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta \).
The clock will lose time i.e. will become slow if θ’ > θ and will gain time i.e. will become fast if θ’ < θ. The gain or loss in time is independent of time period T and depends on the time interval (t). Since, coefficient of linear expansion (α) is very small for invar; hence pendulums are made of invar to show the correct time in all seasons.
]]>The most general form of a quadratic equation x and y is ax² + 2hxy + by² + 2gx + 2fy + c = 0. Since it is an equation in x and y, it must represent the equation of a locus in a plane. It may represent a pair of straight lines, circles, or other curves in different cases. Now, we will consider the case when the above equation represents two straight lines.
Condition for General Second – Degree Equation in x and y Represent a Pair of Straight Lines:
The given equation is
ax² + 2hxy + by² + 2gx + 2fy + c = 0 … (1)
if a ≠ 0, then writing (1) as a quadratic equation in x, we get
ax² + 2x (hy + g) + (by² + 2fy + c) = 0
Solving, we have
\(x=\frac{-2(hy+g)\pm \sqrt{4{{(hy+g)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\).
\(=\frac{-(hy+g)\pm \sqrt{({{h}^{2}}-ab){{y}^{2}}+2(gh-afy)y+({{g}^{2}}-ac)}}{a}\).
Equation (1) will represent two straight lines if the LHS of (i) can be resolved into two linear factors. therefore, the expression under the square roots should be a perfect square.
Hence,
4(gh – af)² – 4(h² – ab)(g² – ac) = 0
(gh – af)² – (h² – ab)(g² – ac) = 0
g²h² + a²f² – 2afgh – h²g² + abg² + ach² – a²bc = 0
a (af² + bg² + ch² – 2fgh – abc) = 0
abc + 2fgh – af² – bg² – ch² = 0
Example: Find the lines whose combined equation is 6x² + 5xy – 4y² + 7x + 13y – 3 = 0
Solution: Given that,
6x² + 5xy – 4y² + 7x + 13y – 3 = 0
6x² + (5y + 7)x – (4y² – 13y + 3) = 0
\(x=\frac{-(5y+7)\pm \sqrt{{{(5y+7)}^{2}}+24(4{{y}^{2}}-13y+3)}}{12}\).
\(x=\frac{-(5y+7)\pm \sqrt{121{{y}^{2}}-242y+121}}{12}\).
\(x=\frac{-(5y+7)\pm 11(y-1)}{12}\).
\(x=\frac{6y-18}{12},\frac{-16y+4}{12}\).
\(x=\frac{y-3}{2},\frac{-4y+1}{3}\).
\(x=\frac{y-3}{2}\).
2x – y + 3 = 0,
\(x=\frac{-4y+1}{3}\).
3x + 4y – 1 = 0.
]]>Dhirubhai Ambani Institute of Information and Communication Technology (DA-IICT), Gandhinagar represents Wave-4 of educational innovation in Gujarat. DA-IICT is spread over 50 acres of land in Gandhinagar, Capital City of Gujarat. The DA-IICT campus is caringly planned and designed as an environmentally conscious campus in the country. The architecture of DA-IICT is functional, but what surrounds it is a fascinating garden. The entire design is oriented towards preserving the environment.
Courses Offered:
⇒ B. Tech (Information and Communication Technology (ICT))
⇒ B. Tech (Honours) in ICT with minor in Computational Science
Eligibility Criteria:
Age:
Only those candidates whose date of birth falls on or after 01 October 1994 are eligible to apply.
Academic Qualification:
Candidate must have passed in final examination of 10+2 (Class XII) or its equivalent referred to as the qualifying examination (mentioned below).
The candidate must have passed in final examination of 10+2 (Class XII) or its equivalent with Mathematics, Physics and any one of Chemistry/ Bio-technology/ Computer Science/ Biology.
Candidates appearing in the qualifying examination in 2019 are also eligible to apply for consideration of provisional admission.
Selection Criteria:
Admission to the B. Tech. programs will be based on the All India Rank of Joint Entrance Examination 2019 (JEE-2019) Main, which is conducted by the National Testing Agency (NTA), Government of India. Candidates who are appearing for JEE 2019 Main can only apply for the programs.
The short-listed candidates will be offered admission (confirmed/waitlisted) in order of their merit (based on the All India Ranking of JEE 2019).
Mode of Application: Online
Application Fee: Rs. 1200/-
Mode of Payment: Online.
Important Dates:
Filling of online Application Forms commences on | 29 April 2019 |
On-line application closes on | 28 June 2019 |
Announcement of First Merit List | 5 July 2019 |
Last Date of Fees Payment for shortlisted candidates | 12 July 2019 |
Registration for Fresh B Tech Students (All India) | 29 July 2019 |
Orientation of Fresh B Tech Students | 31 July – 2 August 2019 |
Commencement of Classes of First year UG Batch and Last Day for Late Registration for Returning Students | 05 August 2019 |
The ratio of the distance from center to focus to distance from center to vertex is called eccentricity.
\(e=\frac{Dis\tan ce\ from\ centre\ to\ focus}{Dis\tan ce\ from\ centre\ to\ vertex}=\frac{c}{a}\).
\({{e}^{2}}=\frac{{{c}^{2}}}{{{a}^{2}}}\).
\({{e}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}=1+\frac{{{b}^{2}}}{{{a}^{2}}}\).
a²e² = a² + b²
Therefore, the equation of hyperbola in terms of eccentricity can be written as
\(\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}({{e}^{2}}-1)}=1\).
The coordinates of the foci are (±ae, 0)
Two hyperbolas are said to be similar if they have the same value of eccentricity
Examples 1: Find the eccentricity of the following hyperbola.
(i) 16y² -9x² = 144
(ii) 5x² – 4y² + 20x + 8y = 4
Solutions:
(i) Given that 16y² -9x² = 144
\(\frac{16{{y}^{2}}}{144}-\frac{9{{x}^{2}}}{144}=1\).
\(\frac{{{y}^{2}}}{144/16}-\frac{{{x}^{2}}}{144/9}=1\).
\(\frac{{{y}^{2}}}{9}-\frac{{{x}^{2}}}{16}=1\) … (1)
\(\frac{{{y}^{2}}}{{{a}^{2}}}-\frac{{{x}^{2}}}{{{b}^{2}}}=1\) … (2)
Compare (1) and (2)
a² = 16 ⇒ a = 4
b² = 9 ⇒ b = 3
eccentricity (e) = a²e² = a² + b²
16e² = 16 + 9
16e² = 25
e² = 25/16 ⇒ e = 5/4
(ii) Given that 5x² – 4y² + 20x + 8y = 4
⇒ 5(x² + 4x) – 4(y² – 2y) = 4
5(x² + 4x -4 + 4) – 4(y² – 2y + 1 – 1) = 4
5(x² + 4x + 4) – 4(y² – 2y + 1) = 4 – 4 + 20
5(x² + 4x + 4) – 4(y² – 2y + 1) = 20
5(x + 2)²- 4(y – 2)² = 20
5(x + 2)² / 20 – 4(y – 2)²/ 20 = 1
(x + 2)²/4 – (y – 2)²/ 5 = 1
a² = 4 ⇒ a = 2
b² = 5
eccentricity (e) = a²e² = a² + b²
= 4 x e² = 4 + 5
e² = 9/4
e = 3/2.
]]>While measuring any physical quantity, it is impossible in principle to find its true value. The difference between the measured and the true values of a physical quantity is called the error of measurement. To determine if a value is accurate compare it to the accepted value. As these values can be anything a concept called percent error has been developed.
What is Percentage Error?
Errors in measurement may also be expressed in terms of percentage. This is similar to relative error except that the error here is converted to a percent value. The percentage error is found by multiplying the relative error by 100%.
It is the absolute value of the difference of the measured value and the actual value divided by the actual value and multiplied by 100.
\(Percentage\,\,Error\,\,=\,\,\frac{Approximate\,Value\,\,-\,\,Exact\,\,Value}{Exact\,\,Value}\times 100\).
How to find the Percentage Error?
Problem:
A scale measures wrongly a value as 10cm due to some marginal errors. Calculate the percentage error if the actual measurement of the value is 15cm?
Solution:
Given,
Exact value = 15cm,
Approximate value = 10cm,
Percentage Error =?
\(Percentage\,\,Error\,\,=\,\,\frac{Approximate\,Value\,\,-\,\,Exact\,\,Value}{Exact\,\,Value}\times 100=\frac{10-15}{15}\times 100\).
\(Percentage\,\,Error\,\,=\frac{10-15}{15}\times 100=-33.33%\).
\(\therefore \,\,Percentage\,\,Error\,\,=-33.33%\).
]]>Tamil Nadu Agricultural University is the premier institution providing agricultural education in Tamil Nadu, with a vision to impart quality education in agriculture and allied sciences and to foster collaboration in agricultural education, research and outreach with national and international institutions of repute. It is the pioneer institute in using innovative and new technologies like ICT enabled e-learning and virtual classrooms.
Eligibility Criteria:
Academic Stream:
Candidates should have passed ALL the subjects in Academic stream of the Qualifying Examination with 10+2 years of schooling and scored minimum eligibility marks as indicated in the information brochure under Board of Higher Secondary Education of Government of Tamil Nadu/ Central Board of Secondary Education/ Council for the Indian school certificate examinations/ other State Government Boards/ other recognized International Boards are eligible to apply for the following degree programs:
Degree | [+2] Subject requirements |
CATEGORY – Ia | |
B.Sc. (Hons) Agriculture | Group I: Mathematics, Physics, Chemistry and Biology OR Group II: Physics, Chemistry, Biology with any one of the fourth (elective) subject viz., Biotechnology, Microbiology, Biochemistry, Computer Science, Home Science. OR Group II (A): Physics, Chemistry, Botany and Zoology |
B.Sc. (Hons) Horticulture | |
B.Sc. (Hons) Forestry | |
B.Sc. (Hons) Sericulture | |
B.Tech. (Biotechnology) | |
B.Sc. (Hons) Agribusiness Management) | |
CATEGORY – Ib | |
B.Sc. (Hons.) Food, Nutrition and Dietetics | Group I: Mathematics, Physics, Chemistry and Biology OR Group II: Physics, Chemistry, Biology with any one of the fourth (elective) subject viz., Biotechnology, Microbiology, Biochemistry, Computer Science, Home Science/ Nutrition and Dietetics OR Group II (A): Physics, Chemistry, Botany and Zoology |
CATEGORY II | |
B.Tech. (Agricultural Engineering) | Group III: Mathematics, Physics, Chemistry and Biology OR Mathematics, Physics, Chemistry and Computer Science |
B.Tech. (Food Technology) | |
B.Tech. (Energy and Environmental Engineering) |
Mode of Selection:
Candidates will be ranked on the basis of their marks in the +2-qualifying examination and called for online counseling based on rank. College and degree allotment will be made based on their rank and preference exercised by the candidate. Separate counseling will be held for special reservations. Certificate verification will be held at Coimbatore campus only. Sliding system will be followed.
Application Fee:
For General & OBC Candidates: Rs. 600/-
For SC, SCA & ST candidates: Rs. 300/-
Mode of Payment: Online
Mode of Application: Online
For Applying: http://www.tnau.ac.in/ugadmission.html
Important Dates:
Opening of online application | 08.05.2019 |
Last date for online application | 07.06.2019 |
Editing/ Correction of online application (3 days only) | 10.06.2019 to 12.06.2019 |
Certificate verification for special reservations (3 days only) (Eminent Sports person, Differently Abled, Ex-Servicemen, Descendants of Freedom Fighter, NRI and Industrial Quota) | 11.06.2019 to 13.06.2019 |
Publication of Rank List | 20.06.2019 |
Rydberg formula is a mathematical formula used to predict the wavelength of light resulting from an electron moving between energy levels of an atom. If the state of an electron in a hydrogen atom is slightly perturbed, then the electron can make a transition to another stationary. The transition will emit a photon with a certain wavelength.
When an electron shifts from an orbital with high energy to a lower energy state, a photon of light is generated. A photon of light gets absorbed by the atom when the electron moves from low energy to a higher energy state. The Rydberg formula is given by:
\(\frac{1}{\lambda }=R{{z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\) ; Where,
\(\lambda \) = Wavelength of the photon,
\(R\) = Rydberg Constant =\(1.097\times {{10}^{7}}{{m}^{-1}}\),
\(Z\) = Atomic number of the atom,
\({{n}_{1}}\) And \({{n}_{2}}\) are integers, where\({{n}_{2}}>{{n}_{1}}\).
How to find the wavelength using Rydberg Formula?
Problem:
Find the wavelength of the electromagnetic radiation that is emitted from an electron relaxes from n=3 to n=1?
Solution: Given,
\(Rydberg\,\,Cons\tan t(R)=1.097\times {{10}^{7}}{{m}^{-1}}\),
\(Z=1\),
\({{n}_{1}}=1\,\,\,\And \,\,\,{{n}_{2}}=3\),
Rydberg formula: \(\frac{1}{\lambda }=R{{z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\),
\(\frac{1}{\lambda }=1.0974\times {{10}^{7}}\left( \frac{1}{{{1}^{1}}}-\frac{1}{{{3}^{2}}} \right)=1.0974\times {{10}^{7}}\left( \frac{1}{1}-\frac{1}{9} \right)\),
\(\frac{1}{\lambda }=1.0974\times {{10}^{7}}(0.889)=0.9755886\times {{10}^{7}}\),
\(\therefore \,\,Wavelength(\lambda )=1.025\times {{10}^{-7}}m.\).
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