Proveriti da li funkcija

\( z = \sin(xy) + y + \mathrm{arctg}\left( \frac{x}{y} \right) \)

zadovoljava sledeći identitet:

\( x^2 \frac{\partial^2 z}{\partial x^2} – y^2 \frac{\partial^2 z}{\partial y^2} + x \frac{\partial z}{\partial x} – y \frac{\partial z}{\partial y} = y \)

Računamo parcijalne izvode:

\( \frac{\partial z}{\partial x} = y \cos(xy) + \frac{y}{x^2 + y^2} \)

\( \frac{\partial^2 z}{\partial x^2} = -y^2 \sin(xy) – \frac{2xy}{(x^2 + y^2)^2} \)

\( \frac{\partial z}{\partial y} = x \cos(xy) + 1 – \frac{x}{x^2 + y^2} \)

\( \frac{\partial^2 z}{\partial y^2} = -x^2 \sin(xy) + \frac{2xy}{(x^2 + y^2)^2} \)

Ubacivanjem u izraz dobijamo:

\[ \begin{aligned} &x^2 \cdot \left(-y^2 \sin(xy) – \frac{2xy}{(x^2 + y^2)^2} \right) – y^2 \cdot \left( -x^2 \sin(xy) + \frac{2xy}{(x^2 + y^2)^2} \right) \\ &+ x \cdot \left( y \cos(xy) + \frac{y}{x^2 + y^2} \right) – y \cdot \left( x \cos(xy) + 1 – \frac{x}{x^2 + y^2} \right) = y \end{aligned} \]

Zaključak: identitet je tačan.

Zamena: \( x + 2 = u \Rightarrow dx = du \)

\( \int \frac{du}{\sqrt{u^2 + 4}} = \ln \left| u + \sqrt{u^2 + 4} \right| + C \)

Rezultat:

\( \boxed{ \int \frac{dx}{\sqrt{x^2 + 4x + 8}} = \ln \left| x + 2 + \sqrt{x^2 + 4x + 8} \right| + C } \)

Zamena: \( u = e^x \Rightarrow dx = \frac{du}{u} \)

\( \int_1^e \frac{1}{u + \frac{4}{u}} \cdot \frac{1}{u} \, du = \int_1^e \frac{1}{u^2 + 4} \, du \)

\( \int \frac{1}{u^2 + 4} du = \frac{1}{2} \mathrm{arctg} \left( \frac{u}{2} \right) \)

Rezultat:

\( \boxed{ \int_0^1 \frac{dx}{e^x + 4e^{-x}} = \frac{1}{2} \left( \mathrm{arctg}\left( \frac{e}{2} \right) – \mathrm{arctg}\left( \frac{1}{2} \right) \right) } \)

\[ 2x y^3 z^3 + x^2 y^3 \cdot 3z^2 \cdot \frac{dz}{dx} + 3x^2 z + x^3 \cdot \frac{dz}{dx} = 0 \]

Grupišemo po \( \frac{dz}{dx} \):

\[ \left( 3x^2 y^3 z^2 + x^3 \right) \cdot \frac{dz}{dx} = – \left( 2x y^3 z^3 + 3x^2 z \right) \]

Rezultat:

\[ \boxed{ \frac{dz}{dx} = – \frac{2x y^3 z^3 + 3x^2 z}{3x^2 y^3 z^2 + x^3} } \]