Question

Find the general solution :​

Answer 1

GIVEN :–

$\\ \bf \implies \dfrac{ { \partial}^{2}z}{ \partial x ^{2} } - 4 \dfrac{ { \partial}^{2}z}{ \partial x \partial y} +4\dfrac{ { \partial}^{2}z}{ \partial y ^{2} } = 0$

TO FIND :–

• General solution = ?

SOLUTION :–

$\\ \bf \implies \dfrac{ { \partial}^{2}z}{ \partial x ^{2} } - 4 \dfrac{ { \partial}^{2}z}{ \partial x \partial y} +4\dfrac{ { \partial}^{2}z}{ \partial y ^{2} } = 0$

• We should write this as –

$\\ \bf \implies{D}^{2}z- 4 DD'z+4 {D'}^{2} z= 0$

• Here –

$\\ \bf \dashrightarrow D \to \dfrac{ \partial}{ \partial x} \: , \: D'\to \dfrac{ \partial}{ \partial y}$

• So that –

$\\ \bf \implies({D}^{2}- 4 DD'+4 {D'}^{2})z= 0$

• Here , R.H.S. is zero . Hence , P.I. = 0

• Now let's find C.F. –

$\\ \bf \implies({D}^{2}- 4 DD'+4 {D'}^{2})z= 0$

• Auxiliary equation –

$\\ \bf \implies{m}^{2}- 4m+4= 0$

$\\ \bf \implies{(m - 2)}^{2}= 0$

$\\ \bf \implies m = 2,2$

• So that C.F. –

$\\ \large\implies \red{ \boxed{ \bf C.F. = \phi_1(y + 2x) +x\phi_2(y + 2x)}}$

• Hence , General solution –

$\\\implies \bf Z =C.F. +P.I.$

$\\\implies \bf Z =\phi_1(y + 2x) +x\phi_2(y + 2x)+0$

$\\ \large\implies \red{ \boxed{ \bf Z= \phi_1(y + 2x) +x\phi_2(y + 2x)}}$