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Find the Differential equation satisfying the family of curves y=ae^(3x)+be^(-2x),a and b are arbitrary constants. |
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Answer» SOLUTION :`y = a e^(3X) + be^(-2X)`, `y_1 = 3ae^(3x) - 2be^(-2x)` `y_2 = 9ae^(3x) + 4be^(-2x)` `= 6ae^(3x) + 6be^(-2x) + 3ae^(3x) - 2be^(-2x)` `= 6Y + y_1` `y_2 - y_1 - 6y = 0`, which i the required differential EQUATION. |
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