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Verify y = a+bx is solution of xd2ydx2+2dydx = 0 y = a+bx dydx=□ d2ydx2=□ Consider xd2ydx2+2dydx = x□+2□ = □ Hence y = a+bx is solution of xd2ydx2+2dydx = 0 - Mathematics and Statistics

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प्रश्न

Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`

रिकाम्या जागा भरा
बेरीज

उत्तर

`x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`("d"y)/("d"x)` = `bb((-b)/x^2)`

`(d^2y)/(dx^2)` = `bb((2b)/x^3)`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `bb(x (2b)/x^3 + 2(-2)/x^2)`

= 0

Hence y = `a + b/x` is solution of `bb(x(d^2y)/(dx^2) + 2(dy)/(dx) = 0)`

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पाठ 1.8: Differential Equation and Applications - Q.6

संबंधित प्रश्‍न

Solve the equation for x: `sin^(-1)  5/x + sin^(-1)  12/x = pi/2, x != 0`


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\[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0\].

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\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]


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