Advertisements
Advertisements
प्रश्न
Show that y = a cos bx is a solution of the! differential equation `("d"^2y)/("d"x^2) + "b"^2y` = 0
Advertisements
उत्तर
Given y = a cos bx .......(1)
Differentiating equation (1) w.r.t ‘x’, we get
`("d"y)/("d"x)` = a(– sin bx)b
`("d"y)/("d"x)` = – ab sin bx
Again differentiating, we get
`("d"^2y)/("d"x^2) = - "ab" cos "b"x * "b"`
y = a cos bx
`y/"a"` = cos bx
`("d"^2y)/("d"x^2) = - "ab"^2 cos "b"x`
`("d"^2y)/("d"x^2) = - "ab"^2 xx y/"a"`
`("d"^2y)/("d"x^2) = - "b"^2y`
`("d"^2y)/("d"x^2)` = 0
Therefore, y = a cos bx is a solution of the differential equation `("d"^2y)/("d"x^2) + "b"^2y` = 0
APPEARS IN
संबंधित प्रश्न
Show the following expressions is a solution of the corresponding given differential equation.
y = aex + be–x ; y'' – y = 0
Find the value of m so that the function y = emx solution of the given differential equation.
y’ + 2y = 0
Find the value of m so that the function y = emx solution of the given differential equation.
y” – 5y’ + 6y = 0
The Slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve
Show that y = e–x + mx + n is a solution of the differential equation `"e"^x(("d"^2y)/("d"x^2)) - 1` = 0
Show that y = `"a"x + "b"/x ≠ 0` is a solution of the differential equation x2yn + xy’ – y = 0
Show that y = ae–3x + b, where a and b are arbitrary constants, is a solution of the differential equation `("d"^2y)/("d"x^2) + 3("d"y)/("d"x)` = 0
Show that the differential equation representing the family of curves y2 = `2"a"(x + "a"^(2/3))`, where a is a postive parameter, is `(y^2 - 2xy ("d"y)/("d"x))^3 = 8(y ("d"y)/("d"x))^5`
Choose the correct alternative:
The general solution of the differential equation `("d"y)/("d"x) = y/x` is
Choose the correct alternative:
The solution of the differential equation `2x ("d"y)/("d"x) - y = 3` represents
Choose the correct alternative:
The solution of the differential equation `("d")/("d"x) + 1/sqrt(1 - x^2) = 0` is
