Math, asked by tusharchauhan89244, 1 month ago

The C.F. of a PDE
022 22
+
дх2 дхду
022
- 6
= x' sin(x+y) is
ду-

Answers

Answered by ashus1550

Step-by-step explanation:

The C.F. of a PDE

022 22

дх2 дхду

022

- 6

= x' sin(x+y) is

ду-The C.F. of a PDE

022 22

дх2 дхду

022

- 6

= x' sin(x+y) is

ду-The C.F. of a PDE

022 22

дх2 дхду

022

- 6

= x' sin(x+y) is

ду-The C.F. of a PDE

022 22

дх2 дхду

022

- 6

= x' sin(x+y) is

ду-The C.F. of a PDE

022 22

дх2 дхду

022

- 6

= x' sin(x+y) is

ду-

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The C.F. of a PDE022 22+дх2 дхду022- 6= x' sin(x+y) isду-​

Answers

The C.F. of a PDE
022 22
+
дх2 дхду
022
- 6
= x' sin(x+y) is
ду-