Differential Equations

f' (x) = x (1+y)
dy/ (y+1) = xdx
ln (y+1) = x²/2 + c
(0,0)
c = 0
y+1 = e^ (x²/2) ⇒ e^ (x²/2) = 2
x²/2 = k = (ln2 > 0)
x = ±√2k

dy/√ (1-y²) = dx/x²
sin? ¹ (y) = -1/x + c ⇒ c = π/2
sin? ¹ (y) = -π/3 + π/2 = π/6
y = 1/2

$y^{2}dx+\left(x^2-xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2-xy+y^2}{y^2}=0\Rightarrow \frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2-\left(\frac{x}{y}\right)+1=0$

Put x = vy $\Rightarrow v+y\frac{dv}{dy}+v^2-v+1=0$

$\Rightarrow \frac{ydv}{dy}+v^2+1=0$

$\Rightarrow \frac{\pi }{6}+\mathcal{l}n\left|y\right|=\frac{\pi }{4}$

$\mathcal{l}n\left|y\right|=\frac{\pi }{12}$

$\left(x+1\right)\frac{dy}{dx}-y=e^{3x}{\left(x+1\right)}^2,$

$\left(x+1\right)dy-ydx=e^{3x}{\left(x+1\right)}^{2}dx$

$\frac{\left(x+1\right)dy-ydx}{{\left(x+1\right)}^2}=e^{3x}dx$

$\Rightarrow d\left(\frac{y}{x+1}\right)=e^{3x}dx\Rightarrow \frac{y}{x+1}=\frac{e^{3x}}{3}+c$

$=\frac{e^{3x}}{3}\left(3x+4\right)$

$\frac{x\left(cosx-sinx\right)}{e^x+1}+\frac{g\left(x\right)\left(e^x+1-x{e}^x\right)}{{\left(e^x+1\right)}^2}$

$=\sqrt[]{2}sin\left(x+\frac{\pi }{4}\right)+c,x>0$

$g\text{'}\left(x\right)=\sqrt[]{2}cos\left(x+\frac{\pi }{4}\right),x>0$

g + g’ = 2cos x + c, x > 0

 g – g’ = 2 sin x + c, x > 0

$f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=x^5+64$

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}\left(\frac{f\left(1\right)}{0}\right)f\left(1\right)=0$

Let f(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e

  f(1) = 0 Þ 1 + a + b + c + d + e = 0

$\Rightarrow x^5+\left(a+5\right)x^4+\left(b+4a+20\right)x^3+\left(c+3b+12a\right)x^2+\left(d+2c+6b\right)x+e+d+2c=x^5+64$  

 Þ e + d + 2c = 64

b + 4a + 20 = 0                 c – 1 – a – b = 64

c + 3b + 12a = 0

d + 2c + 6b = 0   &nb

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$           

$\Rightarrow \frac{dp}{P-900}=0.5dt$         

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$           

T = 2 ln 18

Length of normal
k = y √ (1 + (dy/dx)²)

⇒ ∫ (y dy) / √ (k² - y²) = ∫ dx
Let k² – y² = t²
-2y dy = 2t dt
-√ (k² - y²) = x + c
It passes through (0, k)
x² + y² = k²

. xdy - ydx - x² (xdy + ydx) + 3x? dx = 0
⇒ (xdy - ydx)/x² - (xdy + ydx) + 3x²dx = 0 ⇒ d (y/x) - d (xy) + d (x³) = 0
Integrate both side, we get
y/x - xy + x³ = c
Put x = 3, y = 3
⇒ 1 - 9 + 27 = c
c = 19
Put x = 4
y/4 - 4y = 19 - 64
⇒ y = 12

e? dy = e? /α dx
⇒−e? =e? /α+c
y (ln2)=ln2 and y (0)=−ln2
⇒−2=−1/α+c
⇒c=−2−1/α
⇒e? = 1/α e? −2−1/α
⇒−e? ² = 1/α e? ²−2−1/α
⇒2? ¹=3/α
⇒α=2