Problem 02
Find the Laplace transform of   $\displaystyle \int_0^t (u^2 - u + e^{u}) \, du$.
 

Solution 02
$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
 

Hence,
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{\mathcal{L}(t^2 - t + e^t)}{s}$

$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \mathcal{L}(t^2 - t + e^t)$

Problem 01
Find the Laplace transform of   $\displaystyle \int_0^t \dfrac{\sin u}{u} \, du$   if $\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$.
 

Solution 01
$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
 

Since,
$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$

$F(s) = \arctan \left( \dfrac{1}{s} \right)$
 

Then,

Problem 04
Find the Laplace transform of   $f(t) = t \, \sin t$   using the transform of derivatives.
 

Solution 04
$f(t) = t \, \sin t$       ..........       $f(0) = 0$

$f'(t) = t \, \cos t + \sin t$       ..........       $f'(0) = 0$

$f''(t) = (-t \, \sin t + \cos t) + \cos t = -t \, \sin t + 2\cos t$
 

$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$

Problem 03
Find the Laplace transform of   $f(t) = e^{5t}$   using the transform of derivatives.
 

Solution 03
$f(t) = e^{5t}$       ..........       $f(0) = 1$

$f'(t) = 5e^{5t}$
 

$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

$\mathcal{L} (5e^{5t}) = s \, \mathcal{L} (e^{5t}) - 1$

$1 = s \, \mathcal{L} (e^{5t}) - \mathcal{L} (5e^{5t})$

$s \, \mathcal{L} (e^{5t}) - 5 \, \mathcal{L} (e^{5t}) = 1$

Problem 02
Find the Laplace transform of   $f(t) = \sin^2 t$   using the transform of derivatives.
 

Solution 02
$f(t) = \sin^2 t$       ..........       $f(0) = 0$

$f'(t) = 2\sin t \, \cos t = \sin 2t$
 

$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

$\mathcal{L} (\sin 2t) = s \, \mathcal{L} (\sin^2 t) - 0$

$\dfrac{2}{s^2 + 2^2} = s \, \mathcal{L} (\sin^2 t)$

$s \, \mathcal{L} (\sin^2 t) = \dfrac{2}{s^2 + 4}$

Problem 01
Find the Laplace transform of   $f(t) = t^3$   using the transform of derivatives.
 

Solution 01
$f(t) = t^3$       ..........       $f(0) = 0$

$f'(t) = 3t^2$       ..........       $f'(0) = 0$

$f''(t) = 6t$       ..........       $f''(0) = 0$

$f'''(t) = 6$
 

$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$

Problem 11
$y(x^2 + y^2 - 1) \, dx + x(x^2 + y^2 + 1) \, dy = 0$
 

Solution 11
$y(x^2 + y^2 - 1) \, dx + x(x^2 + y^2 + 1) \, dy = 0$

$y(x^2 + y^2) \, dx - y \, dx + x(x^2 + y^2) \, dy + x \, dy = 0$

$[ \, y(x^2 + y^2) \, dx + x(x^2 + y^2) \, dy \, ] - (y \, dx - x \, dy) = 0$

$(x^2 + y^2)(y \, dx + x \, dy) - (y \, dx - x \, dy) = 0$

$(y \, dx + x \, dy) - \dfrac{y \, dx - x \, dy}{x^2 + y^2} = 0$

$d(xy) - d [ \, \arctan (y/x) \, ] = 0$

$\displaystyle \int d(xy) - \int d[ \, \arctan (y/x) \, ] = 0$

Problem 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

Solution 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

$y^3 \, dx + y \, dx + xy^2 \, dy - x \, dy = 0$

$(xy^2 \, dy + y^3 \, dx) + (y \, dx - x \, dy) = 0$

$y^2(x \, dy + y \, dx) + (y \, dx - x \, dy) = 0$

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

$d(xy) + d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle \int d(xy) + \int d\left( \dfrac{x}{y} \right) = 0$

$xy + \dfrac{x}{y} = c$

$xy^2 + x = cy$