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प्रश्न
The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______
उत्तर
y = 1 – log x
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संबंधित प्रश्न
Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
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\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
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`"x"("dy")/("dx")+"y"=3"x"^2-2`
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| Solution | D.E. |
| y = xn | `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0` |
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
