Differential Equations
Get insights from 323 questions on Differential Equations, answered by students, alumni, and experts. You may also ask and answer any question you like about Differential Equations
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
6 months agoContributor-Level 10
dy/dx + (2? (2? -1)/ (2? ¹ (2? -1) = 0
x, y > 0, y (1) = 1
dy/dx = – (2? (2? -1)/ (2? (2? -1)
∫ (2? -1)/2? dy = –∫ (2? -1)/2? dx
log? (2? -1)/log?2 = – log? (2? -1)/log?2 + log? c/log?2
Taking log of base 2.
∴ y = 2 – log?3
New answer posted
6 months agoContributor-Level 10
I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4
New answer posted
7 months agoContributor-Level 9
dy/dx = 2 (x + 1)
dy = 2 (x + 1)dx.
y = (x + 1)² - c (1)
For y = 0, (x+1)² = c ⇒ x = -1 ± √c.
Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx
A = [cx - (x+1)³/3] from -1-√c to -1+√c
= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)
= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 = (4/3)c√c.
Given A = 4√8 / 3 = 8√2 / 3.
(4/3)c^ (3/2) = 8√2 / 3 ⇒ c^ (3/2) = 2√2 = 2^ (3/2) ⇒ c = 2.
∴ y = (x + 1)² - 2.
∴ y (1) = (1 + 1)² - 2 = 2.
New answer posted
7 months agoContributor-Level 10
Given kx^ (k-1) + k * y^ (k-1) * dy/dx = 0.
dy/dx = - (kx^ (k-1) / (ky^ (k-1) = - (x/y)^ (k-1).
The provided solution has dy/dx + (x/y)^ (k-1) = 0.
It seems to relate to k-1 = -1/3, which implies k = 1 - 1/3 = 2/3.
New answer posted
7 months agoContributor-Level 9
dy/dx + 2y tan (x) = sin (x)
I.F. = e^ (∫2tan (x)dx) = e^ (2ln (sec (x) = sec² (x)
Solution is y sec² (x) = ∫sin (x)sec² (x)dx = ∫sec (x)tan (x)dx
⇒ y sec² (x) = sec (x) + c
y (π/3) = 0 ⇒ 0 * sec² (π/3) = sec (π/3) + c ⇒ 0 = 2 + c ⇒ c = -2.
∴ y = (sec (x) - 2) / sec² (x)
Now let g (t) = (t - 2)/t² = 1/t - 2/t² for |t| ≥ 1.
g' (t) = -1/t² + 4/t³
g' (t) = 0 ⇒ t = 4.
g' (t) = 2/t³ - 12/t? g' (4) < 0, hence maximum.
∴ g (t)max = g (4) = (4 - 2)/4² = 2/16 = 1/8.
New answer posted
7 months agoContributor-Level 10
dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C => C = 0.
So, y * secx = ln (secx).
y = cosx * ln (secx).
At x = π/4:
y = cos (π/4) * ln (sec (π/4) = (1/√2) * ln (√2) = (1/√2) * (1/2)ln (2) = ln (2) / (2√2).
New answer posted
7 months agoContributor-Level 10
The differential equation is rearranged to dt/dx - xt = -e? ²/², where t = 1/ (y+1).
This is a linear first-order differential equation. The integrating factor (I.F.) is e^ (∫-x dx) = e? ²/².
The solution is t * (I.F.) = ∫ Q (x) * (I.F.) dx + c.
t * e? ²/² = ∫ -e? ²/² * e? ²/² dx + c = ∫ -1 dx = -x + c.
Substituting t = 1/ (y+1) back: e? ²/² / (y+1) = -x + c.
Using the initial condition y (2) = 0:
e? ²/ (0+1) = -2 + c ⇒ c = e? ² + 2.
The solution is e? ²/² / (y+1) = 2 + e? ² - x.
New answer posted
7 months agoContributor-Level 10
Given the integral In = ∫(log|x|)^n / x^19 dx.
Let t = log|x|, which implies x = e^t and dx = e^t dt.
The integral becomes:
In = ∫ e^(-20t) * t^n dt
Using integration by parts, where u = t^n and dv = e^(-20t) dt:
In = [t^n * e^(-20t) / -20] - ∫ n*t^(n-1) * e^(-20t) / -20 dt
In = e^(-20) / -20 - (n / -20) * In-1
20 * In = -e^(-20) + n * In-1
For n = 10: 20 * I10 = e^20 - 10 * I9 (Note: There seems to be a sign inconsistency in the original document's derivation vs. standard integration by parts, the document states e^20 instead of -e^(-20) and proceeds with e^20).
For n = 9: 20 * I9 = e^20 - 9 * I8
From these two equations, we can express e
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 66k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers