### Author Topic: TUT0801  (Read 763 times)

#### suyichen

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##### TUT0801
« on: October 04, 2019, 02:54:51 PM »
Show that the DE is not exact and become exact mutiplied by the integrating factor. Then find the solution for the DE
$$( \frac { \operatorname { sin } ( y ) } { y } - 2 e ^ { - x } \operatorname { sin } ( x ) ) + ( \frac { \operatorname { cos } ( y ) + 2 e ^ { - x } \operatorname { cos } ( x ) } { y } ) y ^ { \prime } = 0 , \quad \mu ( x , y ) = y e ^ { x };$$

Let $M _ { ( x , y ) } = \frac { \operatorname { sin } y } { y } - 2 e ^ { - x } \operatorname { sin } x \quad N _ { ( x , y ) } = \frac { \operatorname { cos } y + 2 e ^ { - x } \operatorname { cos } x } { y }$

\qquad$M_y = \frac { y \operatorname { cos } y - \operatorname { sin } y } { y ^ { 2 } } \quad N_x = - \frac { 2 } { y } e ^ { - x } \operatorname { cos } x - \frac { 2 } { y } e ^ { - x } \operatorname { sin } x$

Since  $M_y\neq N_x$ , which the given DE is not exact.

So,multipliying thr integrating fact $\mu(x,y)=ye^{x}$
$$\left. \begin{array} { l } { ( e ^ { x } \operatorname { sin } y - 2 y \operatorname { sin } x ) + ( e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x ) y ^ { \prime } = 0 } \\ { M ^ { \prime } ( x , y ) = e ^ { x } \operatorname { sin } y - 2 y \operatorname { sin } x \quad N ^ { \prime } ( x , y ) = e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x } \\ { M _ { y } ^ { \prime } = e ^ { x } \operatorname { cos } y - 2 \operatorname { sin } x \quad N ^ { \prime }_x = e ^ { x } \operatorname { cos } y - 2 \operatorname { sin } x } \end{array} \right.$$

Since $M^{\prime}y=N^{\prime}x$, which the DE is exact now.

$$\left. \begin{array}{l}{\exists~\psi _ { ( x , y ) } s.t. \psi_y=N^{\prime}=e^x\cos y+2\cos x}\\{ \psi _ { ( x , y ) } = \int N ^ { \prime } d y = \int e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x d y }\\{ \qquad~= e ^ { x } \operatorname { sin } y + 2 y \operatorname { cos } x + h ( x ) }\\{ \psi _ { x } = \operatorname { sin } y e ^ { x } - 2 y \operatorname { sin } x \cdot \operatorname { th } ^ { \prime } ( x ) = M ^ { \prime } }\\{\therefore h^{\prime}(x)=0\quad h(x)=c}\\{\therefore \psi(x,y)=e^x\sin y+2y\cos x}\end{array} \right.$$

Therefore, the solution of the DE is
$$e^x\sin y+2y\cos y=c$$