Exercise 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. y = ex + 1 : y’’ - y’ = 0

Solun:- Given eq. is:- y = ex + 1

Differentiate w.r.t. to x:-

⇒ y’ = ex……....(1)

Again differentiate w.r.t. to x:-

⇒ y’’ = ex……....(2)

Eq. 2 - Eq. 1:-

⇒ y’’ - y’ = 0 this is the given differential equation.

Hence Proved……

2. y = x2 + 2x + C : y’ - 2x - 2 = 0

Solun:- Given eq. is:- y = x2 + 2x + C

We know that:-

Order of differential equation = No. of arbitrary constants

Differentiate w.r.t. to x:-

⇒ y’ = 2x + 2

⇒ y’ - 2x - 2 = 0 this is the given differential equation.

Hence Proved……

3. y = cos x + C : y’ + sin x = 0

Solun:- Given eq. is:- y = cos x + C

We know that:-

Order of differential equation = No. of arbitrary constants

Differentiate w.r.t. to x:-

⇒ y’ = - sin x

⇒ y’ + sin x = 0 this is the given differential equation.

Hence Proved……

${"type":"align","code":"\\begin{align}\n{4.\\,y}&={{\\sqrt[]{1+x^{2}}}}\t\n\\end{align}","font":{"family":"Arial","color":"#000000","size":10},"id":"1-0","ts":1603618312743,"cs":"kS/I0vxjaLfsMk5O3TGCTg==","size":{"width":102,"height":21}}$ : ${"font":{"color":"#434343","size":10,"family":"Arial"},"id":"2","code":"\\begin{align}\n{y^{\\prime }}&={\\frac{xy}{1+x^{2}}}\t\n\\end{align}","type":"align","ts":1603618400825,"cs":"ecnsTPtkiTLPtMo6Oet9rw==","size":{"width":81,"height":29}}$

Solun:- Given eq. is:-

${"id":"1-1","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{y}&={{\\sqrt[]{1+x^{2}}}}\t\n\\end{align*}","ts":1603618428332,"cs":"uQUF/UaInMCxSC22PCstrQ==","size":{"width":88,"height":21}}$

Differentiate w.r.t. to x:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"3-0","code":"\\begin{align*}\n{y^{\\prime }}&={\\frac{1}{2}\\left(1+x^{2}\\right)^{\\frac{-1}{2}}\\times2x}\\\\\n{y^{\\prime }}&={\\frac{x}{\\left(1+x^{2}\\right)^{\\frac{1}{2}}}}\\\\\n{y^{\\prime }}&={\\frac{x}{{\\sqrt[]{1+x^{2}}}}}\t\n\\end{align*}","ts":1603618602874,"cs":"w54ScHvoqwrPYAjXkjtYhg==","size":{"width":156,"height":113}}$

${"code":"\\begin{align*}\n{y^{\\prime }}&={\\frac{x}{{\\sqrt[]{1+x^{2}}}}\\times\\frac{{\\sqrt[]{1+x^{2}}}}{{\\sqrt[]{1+x^{2}}}}}\\\\\n{y^{\\prime }}&={\\frac{x{\\sqrt[]{1+x^{2}}}}{1+x^{2}}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"3-1-0","ts":1604148453426,"cs":"f2NjDW+d2Xy/FRz6cOVo3Q==","size":{"width":178,"height":85}}$

From given equation:- ${"type":"align*","id":"1-0","code":"\\begin{align*}\n{y}&={{\\sqrt[]{1+x^{2}}}}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1604148368773,"cs":"PISLTrGab0F0scwXXzsCsA==","size":{"width":88,"height":21}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{y^{\\prime }}&={\\frac{xy}{1+x^{2}}}\t\n\\end{align*}","id":"3-1-1","ts":1604148470132,"cs":"Kfjn0x43XUFOe/Hl0jwmUA==","size":{"width":81,"height":29}}$ this is the given differential equation.

Hence Proved……

5. y = Ax : xy’ = y (x ≠ 0)

Solun:- Given eq. is:- y = Ax

We know that:-

Order of differential equation = No. of arbitrary constants

Differentiate w.r.t. to x:-

⇒ y’ = A

From given eq.:-

⇒ A = y/x

⇒ y’ = y/x, x ≠ 0

⇒ xy’ = y, (x ≠ 0) this is the given differential equation.

Hence Proved……

6. y = x.sin x : ${"font":{"size":10,"family":"Arial","color":"#434343"},"type":"align","id":"2","code":"\\begin{align}\n{xy^{\\prime }}&={y+x{\\sqrt[]{x^{2}-y^{2}}}}\t\n\\end{align*}","ts":1603619205632,"cs":"l9qflSmVZa06zIpepMlWEQ==","size":{"width":142,"height":20}}$ (x ≠ 0 and x > y or x < -y)

Solun:- Given eq. is:- y = x.sinx

Differentiate w.r.t. to x:-

⇒ y’ = x.cos x + sin x

From given eq.:-

⇒ sin x = y/x and cos x = ${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"${\\sqrt[]{\\frac{x^{2}-y^{2}}{x^{2}}}}$","id":"4","type":"$","ts":1603619500935,"cs":"29sbGJpq4sDIwXClEehaGQ==","size":{"width":52,"height":28}}$ Here x ≠ 0 and x > y or x < -y

${"type":"align*","code":"\\begin{align*}\n{y^{\\prime }}&={x.{\\sqrt[]{\\frac{x^{2}-y^{2}}{x^{2}}}}+\\frac{y}{x}}\\\\\n{y^{\\prime }}&={{\\sqrt[]{x^{2}-y^{2}}}+\\frac{y}{x}}\t\n\\end{align*}","id":"5","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1603619642956,"cs":"UjA4LDFjPEWJn50h+FwXdw==","size":{"width":156,"height":72}}$

${"font":{"family":"Arial","color":"#222222","size":10},"type":"align*","id":"6","code":"\\begin{align*}\n{xy^{\\prime }}&={x{\\sqrt[]{x^{2}-y^{2}}}+y}\t\n\\end{align*}","ts":1603619669235,"cs":"UICh1X6HQCcgh4P8p5l12g==","size":{"width":142,"height":20}}$ , (x ≠ 0 and x > y or x < -y)

This is the given differential equation.

Hence Proved……

7. xy = log y + C : ${"id":"2","font":{"size":10,"color":"#434343","family":"Arial"},"type":"align","code":"\\begin{align}\n{y^{\\prime }}&={\\frac{y^{2}}{1-xy}}\t\n\\end{align*}","ts":1603620024398,"cs":"YohUw45vsuu5eqfqfqSbTw==","size":{"width":84,"height":37}}$ (xy ≠ 1)

Solun:- Given eq. is:- xy = log y + C

Differentiate w.r.t. to x:-

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"7-0","code":"\\begin{align*}\n{xy^{\\prime }+y}&={\\frac{y^{\\prime }}{y}}\t\n\\end{align*}","ts":1603620437323,"cs":"9OnxJWV8BqbUQxmnXlebfw==","size":{"width":88,"height":36}}$

⇒ xyy’ + y2 = y’

⇒ y2 = y’ - xyy’

⇒ y2 = y’(1 - xy)

${"id":"8-0","code":"\\begin{align*}\n{y^{\\prime }}&={\\frac{y^{2}}{1-xy}}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603620563837,"cs":"Rc13ZKfuWKOBQ4GXmrfjNQ==","size":{"width":84,"height":37}}$ , (xy ≠ 1)

This is the given differential equation.

Hence Proved……

8. y - cos y = x : (ysin y + cos y + x).y’ = y

Solun:- Given eq. is:- y - cos y = x

Differentiate w.r.t. to x:-

⇒ y’ + (sin y).y’ = 1

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{y^{\\prime }}&={\\frac{1}{1+\\sin y}}\t\n\\end{align*}","type":"align*","id":"8-1","ts":1603621199416,"cs":"LKVFi41OMbfLFZFt1Dov1g==","size":{"width":96,"height":34}}$

Given differential equation is: (ysin y + cos y + x).y’ = y

Taking L.H.S.

= (ysin y + cos y + x).y’

From given eq.:-

⇒ y = x + cos y

= (ysin y + y).y’

Put the value of y’:-

${"code":"\\begin{align*}\n{}&={\\left(y\\sin y+y\\right).\\left(\\frac{1}{1+\\sin y}\\right)}\\\\\n{}&={y.\\left(\\sin y+1\\right).\\left(\\frac{1}{1+\\sin y}\\right)}\\\\\n{}&={y}\t\n\\end{align*}","id":"9","type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1603621465513,"cs":"zsBWdIef8vFOadH/fSXe/A==","size":{"width":200,"height":100}}$

= R.H.S.

Hence Proved……

9. x + y = tan-1y : y2y’ + y2 + 1 = 0

Solun:- Given eq. is:- x + y = tan-1y

Differentiate w.r.t. to x:-

${"code":"\\begin{align*}\n{1+y^{\\prime }}&={\\frac{y^{\\prime }}{1+y^{2}}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"8-2-0","ts":1603622950889,"cs":"5Ej6WG3RKaTRc8AYR6rHzw==","size":{"width":108,"height":36}}$

⇒ (1 + y’).(1 + y2) = y’

⇒ 1 + y2 + y’ + y’y2 = y’

⇒ y2y’ + y2 + 1 = 0 this is the given differential equation

Hence Proved……

10. ${"font":{"color":"#000000","family":"Arial","size":10},"id":"10","code":"\\begin{align}\n{y}&={{\\sqrt[]{a^{2}-x^{2}}}}\t\n\\end{align}","type":"align","ts":1603623137838,"cs":"CH5liMkNFqd8j7YjkB0sYQ==","size":{"width":94,"height":21}}$ , x ∈ (-a, a) : ${"type":"align","font":{"color":"#000000","size":10,"family":"Arial"},"id":"11","code":"\\begin{align}\n{x+y\\diff{y}{x}}&={0}\t\n\\end{align}","ts":1603623192730,"cs":"VFzgavzid2LQQI5/TXrfVw==","size":{"width":89,"height":32}}$ (y ≠ 0)

Solun:- Given eq. is:-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"10","code":"\\begin{align*}\n{y}&={{\\sqrt[]{a^{2}-x^{2}}}}\t\n\\end{align*}","type":"align*","ts":1603623137838,"cs":"CH5liMkNFqd8j7YjkB0sYQ==","size":{"width":94,"height":21}}$

Differentiate w.r.t. to x:-

${"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{1}{2}\\left(a^{2}-x^{2}\\right)^{\\frac{-1}{2}}\\times\\left(-2x\\right)}\\\\\n{\\diff{y}{x}}&={\\frac{-x}{\\left(a^{2}-x^{2}\\right)^{\\frac{-1}{2}}}}\\\\\n{\\diff{y}{x}}&={\\frac{-x}{{\\sqrt[]{a^{2}-x^{2}}}}}\t\n\\end{align*}","type":"align*","id":"8-2-1","ts":1603623546549,"cs":"HYlRPqQnh6CXQLoF4hPQkg==","size":{"width":201,"height":122}}$

From given eq.:-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"10","code":"\\begin{align}\n{y}&={{\\sqrt[]{a^{2}-x^{2}}}}\t\n\\end{align}","type":"align*","ts":1603623137838,"cs":"CH5liMkNFqd8j7YjkB0sYQ==","size":{"width":94,"height":21}}$

${"type":"align*","id":"8-2-2","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{-x}{y}}\t\n\\end{align*}","ts":1603623664805,"cs":"A0x/TDG6Zjf8GHs7ApP1PA==","size":{"width":73,"height":36}}$

${"id":"12","code":"\\begin{align*}\n{x+y\\diff{y}{x}}&={0}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1603623709471,"cs":"TagyjmU2m+dwvn03A69eWg==","size":{"width":89,"height":32}}$ this is the given differential equation

Hence Proved……

11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Solun:- Given the order of differential equation is 4

We know that:-

Order of differential equation = No. of arbitrary constants

Then no. of arbitrary constants are also 4.

The correct answer is D.

12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Solun:- Given the order of differential equation is 3

But according to the definition of a particular solution:- no. of arbitrary constants are zero because a particular solution is obtained by putting the values to the arbitrary constant in the general solution.

The correct answer is D.

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NCERT Maths solutions for 12th

Important Note

Exercise 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. y = ex + 1 : y’’ - y’ = 0

2. y = x2 + 2x + C : y’ - 2x - 2 = 0

3. y = cos x + C : y’ + sin x = 0

5. y = Ax : xy’ = y (x ≠ 0)

6. y = x.sin x : ${"font":{"size":10,"family":"Arial","color":"#434343"},"type":"align","id":"2","code":"\\begin{align}\n{xy^{\\prime }}&={y+x{\\sqrt[]{x^{2}-y^{2}}}}\t\n\\end{align*}","ts":1603619205632,"cs":"l9qflSmVZa06zIpepMlWEQ==","size":{"width":142,"height":20}}$ (x ≠ 0 and x > y or x < -y)

7. xy = log y + C : ${"id":"2","font":{"size":10,"color":"#434343","family":"Arial"},"type":"align","code":"\\begin{align}\n{y^{\\prime }}&={\\frac{y^{2}}{1-xy}}\t\n\\end{align*}","ts":1603620024398,"cs":"YohUw45vsuu5eqfqfqSbTw==","size":{"width":84,"height":37}}$ (xy ≠ 1)

8. y - cos y = x : (ysin y + cos y + x).y’ = y

9. x + y = tan-1y : y2y’ + y2 + 1 = 0

${"font":{"color":"#000000","family":"Arial","size":10},"id":"10","code":"\\begin{align}\n{y}&={{\\sqrt[]{a^{2}-x^{2}}}}\t\n\\end{align}","type":"align*","ts":1603623137838,"cs":"CH5liMkNFqd8j7YjkB0sYQ==","size":{"width":94,"height":21}}$

11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

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Important Note

Differential Equations(Ex. 9.2)

Exercise 9.2

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. y = ex + 1 : y’’ - y’ = 0

2. y = x2 + 2x + C : y’ - 2x - 2 = 0

3. y = cos x + C : y’ + sin x = 0

:

5. y = Ax : xy’ = y (x ≠ 0)

6. y = x.sin x : (x ≠ 0 and x > y or x < -y)

7. xy = log y + C : (xy ≠ 1)

8. y - cos y = x : (ysin y + cos y + x).y’ = y

9. x + y = tan-1y : y2y’ + y2 + 1 = 0

10. , x ∈ (-a, a) : (y ≠ 0)

11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

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7. xy = log y + C : ${"id":"2","font":{"size":10,"color":"#434343","family":"Arial"},"type":"align","code":"\\begin{align}\n{y^{\\prime }}&={\\frac{y^{2}}{1-xy}}\t\n\\end{align*}","ts":1603620024398,"cs":"YohUw45vsuu5eqfqfqSbTw==","size":{"width":84,"height":37}}$ (xy ≠ 1)