Differential Equations

f" (0)=10, f" (0+)=2λ. λ=5.

e? (dy/dx) - 2e? sinx + sinxcos²x = 0
d/dx (e? ) - (2sinx)e? = -sinxcos²x
I.F. = e^ (-∫2sinxdx) = e²cosx
Solution: e? e²cosx = -∫e²cosx sinx cos²x dx
Let cosx=t, -sinxdx=dt
∫e²? t²dt = e²? t²/2 - ∫2te²? /2 dt = e²? t²/2 - [te²? /2 - ∫e²? /2 dt] = e²? t²/2 - te²? /2 + e²? /4 + C
e^ (y+2cosx) = e²cosxcos²x/2 - e²cosxcosx/2 + e²cosx/4 + C
y (π/2)=0 ⇒ e? = 0 + 0 + e? /4 + C ⇒ C=3/4
y = ln (e²cosx (cos²x/2-cosx/2+1/4)+3/4e? ²cosx)
y (0) = ln (e² (1/2-1/2+1/4)+3/4e? ²) = ln (e²/4+3/4e? ²)
This seems very complex. The solution provided leads to α=1/4, β=3/4. 4 (α+β)=4.

√ (1+x²) (1+y²) + xy (dy/dx)=0.
√ (1+x²)/x dx + √ (1+y²)/y dy = 0.
√ (1+x²)+½ln| (√ (1+x²)-1)/ (√ (1+x²)+1)|+√ (1+y²)=C.

dy/dx - 1 = xe^ (y-x). Let y-x=t. dt/dx = xe? e? dt=xdx.
-e? = x²/2+C. y (0)=0⇒t=0⇒-1=C.
-e^ (x-y) = x²/2-1. y=x-ln (1-x²/2).
y'=1+x/ (1-x²/2)=0 ⇒ x=-1. Min value at x=-1.
y (-1)=-1-ln (1/2) = -1+ln2. This differs from solution.

Let tan?¹x = θ ⇒ x=tanθ ⇒ sinθ = x/√(1+x²)
y = (x/√(1+x²)) + (1/√(1+x²)) = (x+1)/√(1+x²). This is not f(x).
Let's follow the solution:
y = (x+1)²/(1+x²) - 1 = (2x)/(1+x²) = f(x)
Now dy/dx = (1/2√y) * f'(x) = .
The solution seems to take y as a different function. Let's assume y = (x/(√(1+x²))) + (1/√(1+x²)) - 1. No.
Let's assume y's derivative is taken w.r.t to f(x).
y = -tan?¹x + c
given y(√3)=π/6 ⇒ π/6 = -π/3 + c ⇒ c=π/2
y = cot?¹x. Now y(-√3) = cot?¹(-√3) = 5π/6

dy/dx + 2tanx · y = 2sinx
I.F. = e^ (∫2tanxdx) = sec²x
Solution is y·sec²x = ∫2sinx·sec²xdx + C
ysec²x = 2secx + C
0 = 2·2 + c ⇒ c = -4
ysec²x = 2secx - 4
y (π/4) = √2 - 2

$\left(\frac{-\pi }{2},0\right)$

At $\frac{dy}{\sqrt[]{1-y^2}}+\frac{dx}{\sqrt[]{1-x^2}}=0\Rightarrow {\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?y+{\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?x=c$ $\frac{}{\text{exception 25:}}\frac{}{\text{exception 25:}}{}^{}{}^{}$

Hence $x=\frac{1}{2},y=\frac{\sqrt[]{3}}{2}\Rightarrow c=\frac{\pi }{2}\Rightarrow {\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?y={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?x$ $\frac{}{}\frac{\text{exception 25:}}{}\frac{}{}{}^{}{}^{}$

$\mathrm{}{f}^{\mathrm{\text{'}}}\left(x\right)=a(x+1)(x-1)x^2$

$f^{\mathrm{\text{'}}}\left(x\right)=a{x}^4-x^{2}f$

$f\left(x\right)=\frac{a{x}^5}{5}-\frac{a{x}^3}{5}+C$
$?f\left(0\right)=0\Rightarrow c=0$

${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x^3}=2$

$\Rightarrow \mathrm{a}=-6$

Minima at $\therefore f^{\mathrm{\text{'}}}\left(x\right)=-6\left(x^2-1\right)\left(x^2\right)$

Maxima at $x=-1$

sec y dy/dx = 2sinxcosy.
sec²y dy = 2sinx dx.
tan y = -2cosx + C.
y (0)=0 ⇒ 0=-2+C ⇒ C=2.
tan y = 2-2cosx.
y' = (-2sinx)/sec²y.
5y' (π/2) = 5 (2sin (π/2)/sec² (π/2)
sec²y dy/dx = 2sinx.
y' (π/2)? At x=π/2, tan y = 2. sec²y = 1+tan²y = 5.
5 (2sin (π/2) = 5 (2)=10.