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(a) c - k/e^t = p

dp = ke^-t ⇒ p = -ke^-t + c

p = c - k/e^t

(c) y = log(e^x + c) 

dy/dx = e^x/e^y

\(\int\)e^y dy = \(\int\)e^x dx ⇒ e^y = e^x + c

y = log(e^x + c)

(b) 1/x

x(dy/dx) - y = x² ⇒ dy/dx - y/x = x

IF = e^{\(\int\)-1/x dx} = e^-logx = 1/x

The correct answer is : (d) e^∫pdy

The auxiliary equations A.E is m² + 2m + 3 = 0 

⇒ m² + 2m + 1 + 2 = 0 

⇒ (m + 1)² = -2 

⇒ m + 1 = ± √2i 

⇒ m = – 1 ± √2i 

It is of the form α ± iβ 

The complementary function (C.F) = e^-x [A cos √2 x + B sin √2 x] 

The general solution is y = e^-x [A cos √2 x + B sin √2 x]

Given (D² – 2kD + k² )y = 0, D = d/dx

The auxiliary equations is m² – 2km + k = 0

⇒ (m – k)² = 0 

⇒ m = k, k 

Roots are real and equal 

The complementary function (C.F) is (Ax + B) e^kx 

The general solution is y = (Ax + B) e^kx

The highest order derivative is d²y/dx² and its highest power is 2.

So, the degree of differential equation is 2.

\(x^3 \left(\frac{d^2y}{dx^2}\right)^2 + x \left(\frac{dy}{dx}\right)^4=0\)

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is \((\frac{d^2y}{dx^2})^2\)

∴ Degree = 2

\(\left(\frac{dy}{dx}\right)^4+3x\frac{d^2y}{dx^2}=0\)

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is \(\frac{d^2y}{dx^2}\)

∴ Degree = 1

 \(\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2\) \(=x\,sin \left(\frac{dy}{dx}\right)\)

Here Order of differential equation is 2.

Its degree is not defined as it is not a polynomial equation in derivatives.

 \(\frac{d^2y}{dx^2}+3 \left(\frac{dy}{dx}\right)^2\) \(=x^2log \left(\frac{d^2y}{dx^2} \right)\)

Here Order of differential equation is 2.

Its degree is not defined as highest order derivative is a function of logarithmic function and it is not a polynomial equation in derivatives.

\(5x\left(\frac{dy}{dx}\right)^2-\frac{d^2y}{dx^2}-6y=log\,x\)

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is \(\frac{d^2y}{dx^2}\)

∴ Degree = 1

Degree of differential equation is 1 because power of highest order derivative \(\frac{d^y}{dx^2}\) is one.

True, because f ( λx, λy) = λ° f (x, y). 

We have x dy - (y + 2x²) dx = 0

The given differential equation can be written as

⇒ x\(\frac{dy}{dx}\) - y = 2x²

or  \(\frac{dy}{dx}-\frac{1}x.y=2x\)

This is of the form \(\frac{dy}{dx}\) + Py = Q, where P = \(\frac{-1}x,\) Q = 2x

IF = \(e^{-\int\frac{1}{x}dx}=e^{-log\,x}=e^{log\,x^{-1}}=\frac 1x\)

\(\therefore\) Solution is y.\(\frac 1x\) = ∫2x.\(\frac 1x{dx}\)

⇒ y.\(\frac 1x\) = 2x + C

or y = 2x² + Cx

Answer is (D) 4

The number of arbitrary constants in general solution of a differential equation of n^th order is n = 4.

The differential equation is \(\frac{d^2y}{dx^2}+9y = 0\) and the function that is to be proven as the solution is y = 4 sin 3x, now we need to find \(\frac{d^2y}{dx^2}.\)

\(\frac{dy}{dx}\) = 12 cos 3x

\(\frac{d^2y}{dx^2}\) = – 36 sin 3x

Putting the values in the equation, we get,

–36 sin 3x + 9(4 sin 3x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

The differential equation is \(\frac{d^2y}{dx^2}+4y=0\) and the function that is to be proven as solution is

y = A cos 2x – B sin 2x, now we find the value of \(\frac{d^2y}{dx^2}.\)

\(\frac{dy}{dx}\) = = –2A sin 2x – 2B cos 2x

\(\frac{d^2y}{dx^2}\) = –4A cos 2x + 4B sin 2x

Putting the values in the equation, we get,

–4A cos 2x + 4B sin 2x + 4(A cos 2x – B sin 2x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

The differential equation is \(\frac{d^2y}{dx^2}+y = 0\) and the function that is to be proven as solution is y = A cos x + B sin x, now we need to find the value of \(\frac{d^2y}{dx^2}.\)

\(\frac{dy}{dx}\) = –A sin x + B cos x

\(\frac{d^2y}{dx^2}\) = = –A cos x – B sin x

Putting the values in equation, we get,

–A cos x – B sin x + A cos x + B sin x = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

The differential equation is \(\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=0\) and the function that is to be proven as solution is y = ae^2x + be^–x, now we need to find the value of \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}.\)

\(\frac{dy}{dx}\) = 2ae^2x – be^–x

\(\frac{d^2y}{dx^2}\) = = 4ae^2x + be^–x

Putting these values in the equation, we get,

4ae^2x + be^–x –(2ae^2x – be^–x) – 2(ae^2x + be^–x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question, the order of the differential equation is 3, and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is x and the term \(\frac{dx}{dt}\) is multiplied by itself so the given equation is non-linear.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question, the order of the differential equation is 2, and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable y and its derivatives are multiplied with a constant or independent variable only so this equation is linear differential equation.

The given D.E. is \(\cfrac{d^4y}{dx^4}+sin(\frac{dy}{dx})=0\)
This D.E. has highest order derivative \(\cfrac{d^4y}{dx^2}\) . 

∴ order = 4

Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question, we first need to remove the term \(\frac{1}{dy/dx}\) because this can be written as \((\frac{dy}{dx})^{-1}\) which means a negative power.

So, the above equation becomes as

\((\frac{dy}{dx})^3+1 = 2 \frac{dy}{dx}\)

So, in this, the order of the differential equation is 3, and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term \(\frac{dy}{dx}\) is multiplied by itself so the given equation is non-linear.

The given D.E. is \(\cfrac{dy}{dx^2}+X\left(\cfrac{dy}{dx}\right)+y = 2 sin \,x\)
This D.E. has highest order derivative \(\cfrac{d^2y}{dx^2}\) with power 1.

∴ the given D.E. is of order 2 and degree 1.

The given D.E. is \(\cfrac{dy}{dx} =\cfrac{2\, sin\,x+3}{\frac{dy}{dx}}\)
\(\therefore \) \(\left(\cfrac{dy}{dx}\right)^2\)= 2 sin x + 3

This D.E. has highest order derivative dy/dx with power 2.

∴ the given D.E. is of order 1 and degree 2.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question the order of the differential equation is 2 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term \(\frac{dy}{dx}\)is multiplied by itself so the given equation is non-linear.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question, the order of the differential equation is 1, and the degree of the differential equation is 3.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term \(\frac{dy}{dx}\) is multiplied by itself so the given equation is non-linear.

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

Concept of the question

For the degree to be defined of any differential equation the euqtion must be expressible in the form of a polynomial.

But, in this question the degree of the differential equation is not defined because the term on the right hand side is not expressible in the form of a polynomial.

Thus, the order of the above equation is 2 whereas the degree is not defined.

Since the degree of the equation is not defined the equation is non-linear.

y¹¹ + a²y = cos ax

It's complementary equation is:

m² + a² = 0

⇒ m = \(\pm ac\) 

\(\therefore\) C. F. = C₁ cos ax + C₂ sin ax

P.I. = \(\frac{1}{D^2+a^2}cos\,ax=\frac{x}{2a}sin\,ax\) 

\(\therefore\) Complete solution of given differential equation is

y = C.F. + P.I.

 = C₁ cos ax + C₂ sin ax + x/2a sin ax

Given differential equation is

\(\frac{d^2y}{dt}^2 + 6 \frac{dy}{dt} + 8y = 5 sin\,3t\)

\(\Rightarrow\) D²y + 6Dy + 8y = 5 sin 3t (\(\because \frac{d}{dt} = D\) & \(\frac{d^2}{dt^2} = D^2\))

\(\Rightarrow\) (D² + 6D + 8)y = 5sin 3t

The characterstic equation of given differential equation is

\(\Rightarrow\) m² + 6m + 8

\(\Rightarrow\) m² + 2m + 4m + 8 = 0

\(\Rightarrow\) m(m + 2) + 4(m + 4) = 0

\(\Rightarrow\) (m + 2)(m + 4) = 0

\(\Rightarrow\) m = -2, -4

The complementary function is

C. F = C₁e^-2t + C₂e^-4t

Now, particular integral is

P.I = \(\frac{1}{D^2 + 6D + 8} \) 5 sin 3t

= \(\frac{1}{-9 + 6D + 8}\) 5 sin 3t (\(\frac{1}{f(D)^2} = \frac{1}{f(-a)^2} sin ax; \, f(-a^2) \neq 0\))

= \(\frac{6D + 1}{(6D - 1)(6D + 1)}\) 5sin 3t

= \(\frac{6D + 1}{36D^2 - 1}\) 5 sin 3t

= 5 (6D + 1) \(\frac{sin \,3t}{-36 \times 9-1}\)

= \(\frac{-5}{325}\) (6D + 1) sin 3t

= \(-\frac{1}{65} (18 cos\, 3t + sin \, 3t)\)

Hence the solution is y = C.F + P.I

 \(\Rightarrow\) y = C₁e^-2t + C₂e^-4t - \(\frac{1}{65}\)(18 cos 3t + sin 3t)

 Given differential equation is

\(\frac{d^2y}{dt}^2 + 6 \frac{dy}{dt} + 8y = 5 sin\,3t\)

\(\Rightarrow\) D²y + 6Dy + 8y = 5 sin 3t (\(\because \frac{d}{dt} = D\) & \(\frac{d^2}{dt^2} = D^2\))

\(\Rightarrow\) (D² + 6D + 8)y = 5sin 3t

The characterstic equation of given differential equation is

\(\Rightarrow\) m² + 6m + 8

\(\Rightarrow\) m² + 2m + 4m + 8 = 0

\(\Rightarrow\) m(m + 2) + 4(m + 4) = 0

\(\Rightarrow\) (m + 2)(m + 4) = 0

\(\Rightarrow\) m = -2, -4

The complementary function is

C. F = C₁e^-2t + C₂e^-4t

Now, particular integral is

P.I = \(\frac{1}{D^2 + 6D + 8} \) 5 sin 3t

= \(\frac{1}{-9 + 6D + 8}\) 5 sin 3t (\(\frac{1}{f(D)^2} = \frac{1}{f(-a)^2} sin ax; \, f(-a^2) \neq 0\))

= \(\frac{6D + 1}{(6D - 1)(6D + 1)}\) 5sin 3t

= \(\frac{6D + 1}{36D^2 - 1}\) 5 sin 3t

= 5 (6D + 1) \(\frac{sin \,3t}{-36 \times 9-1}\)

= \(\frac{-5}{325}\) (6D + 1) sin 3t

= \(-\frac{1}{65} (18 cos\, 3t + sin \, 3t)\)

Hence the solution is y = C.F + P.I

 \(\Rightarrow\) y = C₁e^-2t + C₂e^-4t - \(\frac{1}{65}\)(18 cos 3t + sin 3t)

1. Order = 1, Degree = 1

2. Given,

 \(\frac{dy}{dx}\) + y = sin x is of the form

\(\frac{dy}{dx}\)+ Py = Q ⇒ P = 1, Q = sinx

Integrating factor = e^∫Pdx = e^∫1dx = e^x

3. Therefore solution is

y.IF = ∫Q.IFdx + c ⇒ ye^x = ∫e^x sinxdx + c ____(1)

∫sinx.e^xdx = e^x sinx – ∫cosx.e^xdx

= e^x sin x – cosx.e^x – ∫sinx.e^x dx

⇒ 2∫e^x sin xdx = e^x(sin x – cos x)

⇒ ∫e^x sinxdx =\(\frac{e^x}{2}\)(sinx – cosx)

(1) ⇒ ye^x = \(\frac{e^x}{2}\)(sinx – cosx) + c.

Correct answer is (C).

Correct answer is (B). Let the equation of given family be (x – h)² + (y – k)² = a² . It has two orbitrary constants h and k. Threrefore, the order of the given differential equation will be 2.

Correct answer is (B).

Correct answer is (C).

 From the given equation, we get logx + logy = logc giving xy = c.

The number of arbitrary constants in a particular solution of the differential equation tan x dx + tan y dy = 0 is __________ .

One; a is the only arbitrary constant.

(a) 1

Since the power of d⁴y/dx⁴ is 1

(c) 2 and 1 

Squaring both sides, we get d²y/dx² = dy/dx + 5

So order = 2, degree = 1

(a) d²y/dx² - y = 0

y = ae^x + be^-x; y' = ae^x - be^-x

y" = ae^x + be^-x ⇒ y" - y' = 0

(a) y = x(dy/dx) + dy/dx - (dy/dx)³

(c) xy = c

x dy + y dx = 0 

d(xy) = 0 

xy = c

(a) x² + y² = c 

x dx = -y dy 

x²/2 = -y²/2 + c₁

x² + y² = c

(i) (d) Not defined.

(ii) \(\frac{dy}{dx}\) + 2y tanx = sinx

Then, P = 2tanx, Q = sinx

IF = e^∫Pdx = e^∫2tanxdx = e^{2log sec x} = sec² x 

Solution is; y × IF = ∫Q(IF)dx + c

⇒ ysec² x = ∫sinx sec² xdx + c

⇒ ysec² x = ∫tanx secx dx + c

⇒ ysec²x = secx + c

Here; y = 0, x = \(\frac{\pi}{3}\)

⇒ 0 × sec² \(\frac{\pi}{3}\)= sec \(\frac{\pi}{3}\) + c ⇒ c = -2

⇒ ysec² x = secx – 2.

(D² + 9)y = sin3x

It's auxiliarly equation is 

m² + 9 = 0

⇒ m = \(\pm\) 3c

\(\therefore\) C.F. = c₁ cos3x + c₂ sin3x

P.I. = \(\frac1{D^2 + 9} sin3x\)

\( = \frac{-x}{2\times3} cos3x\)         \(\left(\because \frac1{D^2 + a^2}sin\,ax = \frac{-x\,cos \,ax}{2a}\right)\) 

\( = \frac{-x}{6}cos3x\)

Particular integral = \( \frac{-x}{6}cos3x\)

The order of a differential equation is the order of the highest derivative involved in the equation. So the 

order comes out to be 2 as we have \(\cfrac{d^2y}{dx^2}\) and the degree is the highest power to which a derivative is raised. So the power at this order is 1. 

So the answer is 2, 1.

The auxiliary equations is m² – 6m + 25 = 0

\(m = {-6 \pm \sqrt{36-100} \over 2}\) 

= (6 \(\pm\) 8i)/2 

= 3 \(\pm\)4i

The Roots are complex and of the form, 

α ± β with α = 3 and β = 4 

The complementary function = e^3x (A cos 4x + B sin 4x) 

The general solution is y = e^3x (A cos 4x + B sin 4x)

The order of a differential equation is the order of the highest derivative involved in the equation. So, the order comes out to be 1 as we have \((3x+5y)dy-4x^2dx=0\)

and the degree is the highest power to which a derivative is raised. So the power at this order is 1. 

So the answer is 1, 1.