Question

solve (D²-18D -77)y=0​

Answer 1

GIVEN :–

• Differentiate equation –

$\\ \implies\bf (D^2-18D -77)y=0$

TO FIND :–

• Solution of differential equation = ?

SOLUTION :–

• We know that –

$\\ \large\implies{ \boxed{\bf Solution=C.F. + P.I.}}$

$\\ \: \: \because\bf \: \:Here \: \:P.I. = 0$

$\\ \: \: \therefore \: \: \large { \boxed{\bf \:Solution=C.F.}}$

• Now let's find C.F. –

$\\ \implies\bf (D^2-18D -77)y=0$

• Auxiliary equation –

$\\ \implies\bf {m}^{2} -18m -77 = 0$

$\\ \implies\bf m= \dfrac{ - ( - 18) \pm \sqrt{ {( -18)}^{2} - 4(1)(-77)} }{2(1)}$

$\\ \implies\bf m= \dfrac{18 \pm \sqrt{324 + 308} }{2}$

$\\ \implies\bf m= \dfrac{18 \pm \sqrt{632} }{2}$

$\\ \implies\bf m= \dfrac{18 \pm 2\sqrt{158} }{2}$

$\\ \implies \large{ \boxed{\bf m=9 \pm \sqrt{158}}}$

$\\ \implies \large{ \boxed{\bf m=9 + \sqrt{158},9-\sqrt{158}}}$

• Hence , C.F. –

$\\ \implies \bf \: C.F. =c_{1} {e}^{(9 + \sqrt{158})}+c_{2} {e}^{(9 - \sqrt{158})}$

• And –

$\\ \implies\large { \boxed{\bf \:Solution=c_{1} {e}^{(9 + \sqrt{158})}+c_{2} {e}^{(9 - \sqrt{158})}}}$

Answer 2

GIVEN :–

• Differentiate equation –

⟹(D

2

−18D−77)y=0

TO FIND :–

• Solution of differential equation = ?

SOLUTION :–

• We know that –

\begin{gathered} \\ \large\implies{ \boxed{\bf Solution=C.F. + P.I.}} \\ \end{gathered}

⟹

Solution=C.F.+P.I.

\begin{gathered} \\ \: \: \because\bf \: \:Here \: \:P.I. = 0\\ \end{gathered}

∵HereP.I.=0

\begin{gathered} \\ \: \: \therefore \: \: \large { \boxed{\bf \:Solution=C.F.}}\\ \end{gathered}

∴

Solution=C.F.

• Now let's find C.F. –

\begin{gathered} \\ \implies\bf (D^2-18D -77)y=0 \\ \end{gathered}

⟹(D

2

−18D−77)y=0

• Auxiliary equation –

\begin{gathered} \\ \implies\bf {m}^{2} -18m -77 = 0\\ \end{gathered}

⟹m

2

−18m−77=0

\begin{gathered} \\ \implies\bf m= \dfrac{ - ( - 18) \pm \sqrt{ {( -18)}^{2} - 4(1)(-77)} }{2(1)} \\ \end{gathered}

⟹m=

2(1)

−(−18)±

(−18)

2

−4(1)(−77)

\begin{gathered} \\ \implies\bf m= \dfrac{18 \pm \sqrt{324 + 308} }{2} \\ \end{gathered}

⟹m=

2

18±

324+308

\begin{gathered} \\ \implies\bf m= \dfrac{18 \pm \sqrt{632} }{2} \\ \end{gathered}

⟹m=

2

18±

632

\begin{gathered} \\ \implies\bf m= \dfrac{18 \pm 2\sqrt{158} }{2} \\ \end{gathered}

⟹m=

2

18±2

158

\begin{gathered} \\ \implies \large{ \boxed{\bf m=9 \pm \sqrt{158}}}\\ \end{gathered}

⟹

m=9±

158

\begin{gathered} \\ \implies \large{ \boxed{\bf m=9 + \sqrt{158},9-\sqrt{158}}}\\ \end{gathered}

⟹

m=9+

158

,9−

158

• Hence , C.F. –

\begin{gathered} \\ \implies \bf \: C.F. =c_{1} {e}^{(9 + \sqrt{158})}+c_{2} {e}^{(9 - \sqrt{158})}\\ \end{gathered}

⟹C.F.=c

1

e

(9+

158

)

+c

2

e

(9−

158

)

• And –

\begin{gathered} \\ \implies\large { \boxed{\bf \:Solution=c_{1} {e}^{(9 + \sqrt{158})}+c_{2} {e}^{(9 - \sqrt{158})}}}\\ \end{gathered}

⟹

Solution=c

1

e

(9+

158

)

+c

2

e

(9−

158

)