EXERCISE 1
Direction: Determine the order and degree (if defined) of the following differential equations:
Q1
\( x \frac{dy}{dx} + 2y = x^2, \quad x \neq 0 \)
Answer: Order: 1, Degree: 1
Reasoning:
- Order: The highest order derivative present in the equation is \( \frac{dy}{dx} \), so the order is 1.
- Degree: The differential equation is a polynomial equation in derivatives. The highest order derivative \( \frac{dy}{dx} \) has the power 1, so the degree is 1.
Q2
\( \frac{dy}{dx} + e^y = 0 \)
Answer: Order: 1, Degree: 1
Reasoning:
- Order: The highest order derivative present is \( \frac{dy}{dx} \), so the order is 1.
- Degree: The equation is a polynomial in its derivatives (the term \( e^y \) involves the dependent variable \( y \), not the derivative). The power of \( \frac{dy}{dx} \) is 1, so the degree is 1.
Q3
\( \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0 \)
Answer: Order: 2, Degree: 1
Reasoning:
- Order: The highest order derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2.
- Degree: The highest order derivative \( \frac{d^2y}{dx^2} \) has a power of 1, so the degree is 1.
Q4
\( \left(\frac{dy}{dx}\right)^4 + 3y \left(\frac{d^2y}{dx^2}\right) = 0 \)
Answer: Order: 2, Degree: 1
Reasoning:
- Order: The highest order derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2.
- Degree: The degree is determined by the power of the highest order derivative. Here, the highest order derivative \( \frac{d^2y}{dx^2} \) has a power of 1 (even though \( \frac{dy}{dx} \) has a higher power of 4). Therefore, the degree is 1.
Q5
\( (y''')^2 + (y'')^3 + (y')^4 + y^5 = 0 \)
where \( y' = \frac{dy}{dx}, y'' = \frac{d^2y}{dx^2} \) and \( y''' = \frac{d^3y}{dx^3} \)
where \( y' = \frac{dy}{dx}, y'' = \frac{d^2y}{dx^2} \) and \( y''' = \frac{d^3y}{dx^3} \)
Answer: Order: 3, Degree: 2
Reasoning:
- Order: The highest order derivative present is \( y''' \) (the third derivative), so the order is 3.
- Degree: The highest order derivative \( y''' \) is raised to the power of 2. Therefore, the degree is 2.
EXERCISE 2
Direction: Verify that given function (explicit or implicit) is a solution of the corresponding differential equation (Q1 to 6)
Q1
\( y = ae^{-x} \quad : \quad \frac{dy}{dx} + y = 0 \)
Verification:
Given function: \( y = ae^{-x} \)
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = a(-e^{-x}) = -ae^{-x} \]Substitute \( \frac{dy}{dx} \) and \( y \) into the L.H.S of the differential equation:
\[ \text{L.H.S} = \frac{dy}{dx} + y \] \[ = (-ae^{-x}) + (ae^{-x}) \] \[ = 0 = \text{R.H.S} \]Hence, the given function is a solution of the differential equation.
Q2
\( y = \sqrt{1+x^2} \quad : \quad \frac{dy}{dx} = \frac{xy}{1+x^2} \)
Verification:
Given function: \( y = (1+x^2)^{1/2} \)
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-1/2} \cdot \frac{d}{dx}(1+x^2) \] \[ \frac{dy}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}} \]Now, substitute \( y = \sqrt{1+x^2} \) into the R.H.S of the differential equation:
\[ \text{R.H.S} = \frac{xy}{1+x^2} = \frac{x(\sqrt{1+x^2})}{1+x^2} \] \[ = \frac{x}{\sqrt{1+x^2}} = \text{L.H.S} \]Hence Verified.
Q3
\( xy = \log y + c \quad : \quad \frac{dy}{dx} = \frac{y^2}{1-xy}, \quad (xy \neq 1) \)
Verification:
Given: \( xy = \log y + c \)
Differentiating both sides with respect to \( x \):
\[ x\frac{dy}{dx} + y(1) = \frac{1}{y}\frac{dy}{dx} \] \[ y = \frac{1}{y}\frac{dy}{dx} - x\frac{dy}{dx} \] \[ y = \frac{dy}{dx} \left( \frac{1}{y} - x \right) \] \[ y = \frac{dy}{dx} \left( \frac{1-xy}{y} \right) \] \[ \frac{dy}{dx} = \frac{y^2}{1-xy} \]This matches the given differential equation. Hence Verified.
Q4
\( ax^2 + by^2 = 1 \quad : \quad x(yy_2 + y_1^2) = yy_1 \)
where \( y_1 = \frac{dy}{dx}, \quad y_2 = \frac{d^2y}{dx^2} \)
where \( y_1 = \frac{dy}{dx}, \quad y_2 = \frac{d^2y}{dx^2} \)
Verification:
Differentiating \( ax^2 + by^2 = 1 \) with respect to \( x \):
\[ 2ax + 2byy_1 = 0 \Rightarrow ax + byy_1 = 0 \quad \dots(1) \]Differentiating again with respect to \( x \):
\[ a + b(y \cdot y_2 + y_1 \cdot y_1) = 0 \] \[ a + b(yy_2 + y_1^2) = 0 \Rightarrow a = -b(yy_2 + y_1^2) \]Substitute value of \( a \) in equation (1):
\[ [-b(yy_2 + y_1^2)]x + byy_1 = 0 \]Divide by \( -b \):
\[ x(yy_2 + y_1^2) - yy_1 = 0 \] \[ x(yy_2 + y_1^2) = yy_1 \]Hence Verified.
Q5
\( y = (a + bx)e^{2x} \quad : \quad y_2 - 4y_1 + 4y = 0 \)
Verification:
Given: \( y = (a + bx)e^{2x} \)
Differentiating w.r.t \( x \):
\[ y_1 = (a + bx)(2e^{2x}) + e^{2x}(b) \] \[ y_1 = 2y + be^{2x} \quad \dots(1) \]Differentiating again w.r.t \( x \):
\[ y_2 = 2y_1 + b(2e^{2x}) = 2y_1 + 2be^{2x} \]From (1), we know \( be^{2x} = y_1 - 2y \). Substitute this into \( y_2 \):
\[ y_2 = 2y_1 + 2(y_1 - 2y) \] \[ y_2 = 2y_1 + 2y_1 - 4y \] \[ y_2 = 4y_1 - 4y \] \[ y_2 - 4y_1 + 4y = 0 \]Hence Verified.
Q6
\( x^2 = 2y^2 \log y \quad : \quad (x^2 + y^2)\frac{dy}{dx} - xy = 0 \)
Verification:
Differentiating \( x^2 = 2y^2 \log y \) w.r.t \( x \):
\[ 2x = 2 \left[ y^2 \cdot \frac{1}{y} \frac{dy}{dx} + \log y \cdot 2y \frac{dy}{dx} \right] \] \[ x = \left[ y + 2y \log y \right] \frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{x}{y(1 + 2\log y)} \]From the original equation, \( \log y = \frac{x^2}{2y^2} \). Substitute this:
\[ \frac{dy}{dx} = \frac{x}{y \left( 1 + 2 \cdot \frac{x^2}{2y^2} \right)} = \frac{x}{y \left( \frac{y^2 + x^2}{y^2} \right)} \] \[ \frac{dy}{dx} = \frac{x \cdot y^2}{y(x^2 + y^2)} = \frac{xy}{x^2 + y^2} \] \[ (x^2 + y^2)\frac{dy}{dx} = xy \] \[ (x^2 + y^2)\frac{dy}{dx} - xy = 0 \]Hence Verified.
Q7
Verify that the function, \( y = ke^x - 1 \) is a solution of the differential equation \( \frac{dy}{dx} = y + 1 \). Also determine the value of the constant \( k \) so that the solution curve of the given differential equation passes through the point (0,1).
Part 1: Verification
Given \( y = ke^x - 1 \).
L.H.S: \( \frac{dy}{dx} = k(e^x) - 0 = ke^x \)
R.H.S: \( y + 1 = (ke^x - 1) + 1 = ke^x \)
Since L.H.S = R.H.S, the function is a solution.
Part 2: Value of k
The curve passes through the point \( (0, 1) \). Substitute \( x = 0 \) and \( y = 1 \) into the equation:
\[ 1 = ke^0 - 1 \] \[ 1 = k(1) - 1 \] \[ k = 2 \]So, the value of the constant \( k \) is 2.
EXERCISE 3
Q1
Form the differential equation not containing the arbitrary constants and satisfied by the equation \( x^2 - y^2 = a^2 \), where \( a \) is an arbitrary constant.
Answer: \( x - y \frac{dy}{dx} = 0 \)
Given equation: \( x^2 - y^2 = a^2 \)
Differentiating both sides with respect to \( x \):
\[ 2x - 2y \frac{dy}{dx} = 0 \]Dividing by 2:
\[ x - y \frac{dy}{dx} = 0 \]This differential equation is free from the arbitrary constant \( a \).
Q2
Find the differential equation of the family of circles having centre at origin.
Answer: \( x + y \frac{dy}{dx} = 0 \)
The equation of the family of circles with center at origin \((0,0)\) and radius \( r \) is:
\[ x^2 + y^2 = r^2 \]Here, \( r^2 \) is the arbitrary constant. Differentiating with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} = 0 \] \[ 2(x + y \frac{dy}{dx}) = 0 \] \[ x + y \frac{dy}{dx} = 0 \]Q3
Form the differential equation of the family of circles having centre on \( y - axis \) and passing through origin.
Answer: \( (x^2 - y^2) \frac{dy}{dx} - 2xy = 0 \)
Let the center be \((0, a)\). Since the circle passes through the origin \((0,0)\), the radius is \( a \).
The equation is: \( (x - 0)^2 + (y - a)^2 = a^2 \)
\[ x^2 + y^2 - 2ay + a^2 = a^2 \] \[ x^2 + y^2 = 2ay \quad \dots(1) \]Differentiating w.r.t \( x \):
\[ 2x + 2y \frac{dy}{dx} = 2a \frac{dy}{dx} \]From (1), substitute \( 2a = \frac{x^2 + y^2}{y} \) into the differential equation:
\[ 2x + 2y \frac{dy}{dx} = \left( \frac{x^2 + y^2}{y} \right) \frac{dy}{dx} \]Multiply by \( y \):
\[ 2xy + 2y^2 \frac{dy}{dx} = (x^2 + y^2) \frac{dy}{dx} \] \[ 2xy = (x^2 + y^2 - 2y^2) \frac{dy}{dx} \] \[ 2xy = (x^2 - y^2) \frac{dy}{dx} \] \[ (x^2 - y^2)\frac{dy}{dx} - 2xy = 0 \]Q4
Form the differential equation representing the family of curves \( y = e^{2x}(a + bx) \), where \( a, b \) are arbitrary constants.
Answer: \( \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0 \)
Given: \( y = e^{2x}(a + bx) \)
Differentiating w.r.t \( x \):
\[ y' = 2e^{2x}(a + bx) + e^{2x}(b) \]Since \( e^{2x}(a+bx) = y \):
\[ y' = 2y + be^{2x} \quad \dots(1) \]Differentiating again:
\[ y'' = 2y' + 2be^{2x} \]From (1), \( be^{2x} = y' - 2y \). Substitute this into the \( y'' \) equation:
\[ y'' = 2y' + 2(y' - 2y) \] \[ y'' = 2y' + 2y' - 4y \] \[ y'' - 4y' + 4y = 0 \]Q5
Find the differential equation representing the parabolas having their vertices at origin and foci on positive direction of \( x-axis \).
Answer: \( 2x \frac{dy}{dx} - y = 0 \)
The equation of such parabolas is \( y^2 = 4ax \), where \( a \) is an arbitrary constant.
Differentiating w.r.t \( x \):
\[ 2y \frac{dy}{dx} = 4a \]From the parabola equation, \( 4a = \frac{y^2}{x} \). Substitute this:
\[ 2y \frac{dy}{dx} = \frac{y^2}{x} \]Dividing by \( y \) (assuming \( y \neq 0 \)):
\[ 2 \frac{dy}{dx} = \frac{y}{x} \] \[ 2x \frac{dy}{dx} - y = 0 \]Q6
Form the differential equation of the family of ellipses having their foci on \( x - axis \) and centre at the origin.
Answer: \( xy \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 - y \frac{dy}{dx} = 0 \)
Standard equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Differentiating w.r.t \( x \):
\[ \frac{2x}{a^2} + \frac{2y y'}{b^2} = 0 \Rightarrow \frac{x}{a^2} + \frac{y y'}{b^2} = 0 \] \[ \frac{y y'}{x} = -\frac{b^2}{a^2} \]Differentiating again w.r.t \( x \) (using quotient rule on LHS, constant on RHS becomes 0):
\[ \frac{x \frac{d}{dx}(yy') - yy'(1)}{x^2} = 0 \] \[ x(y y'' + (y')^2) - yy' = 0 \] \[ xy \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 - y \frac{dy}{dx} = 0 \]EXERCISE 4
Direction: Find the general solution of the following differential equations given in Q.1 to 5.
Q1
\( \frac{dy}{dx} = (e^x + 1)y \)
Answer: \( \log|y| = e^x + x + C \)
Separating variables:
\[ \frac{dy}{y} = (e^x + 1) dx \]Integrating both sides:
\[ \int \frac{dy}{y} = \int (e^x + 1) dx \] \[ \log|y| = e^x + x + C \]Q2
\( x^5 \frac{dy}{dx} = -y^5 \)
Answer: \( x^{-4} + y^{-4} = C \)
Separating variables:
\[ \frac{dy}{y^5} = -\frac{dx}{x^5} \] \[ y^{-5} dy = -x^{-5} dx \]Integrating both sides:
\[ \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + K \]Multiplying by -4:
\[ y^{-4} = -x^{-4} + C \] \[ x^{-4} + y^{-4} = C \]Q3
\( \frac{dy}{dx} = \frac{x+1}{2-y} \)
Answer: \( 4y - y^2 = x^2 + 2x + C \)
Separating variables:
\[ (2-y) dy = (x+1) dx \]Integrating both sides:
\[ \int (2-y) dy = \int (x+1) dx \] \[ 2y - \frac{y^2}{2} = \frac{x^2}{2} + x + K \]Multiplying by 2:
\[ 4y - y^2 = x^2 + 2x + C \]Q4
\( x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0 \)
Answer: \( e^y + e^{-y} = \log|x| - \frac{x^2}{2} + C \)
Rearranging terms:
\[ x(e^{2y} - 1)dy = -(x^2 - 1)e^y dx \]Separating variables:
\[ \frac{e^{2y} - 1}{e^y} dy = -\frac{x^2 - 1}{x} dx \] \[ (e^y - e^{-y}) dy = (-x + \frac{1}{x}) dx \]Integrating both sides:
\[ \int (e^y - e^{-y}) dy = \int (\frac{1}{x} - x) dx \] \[ e^y - (-e^{-y}) = \log|x| - \frac{x^2}{2} + C \] \[ e^y + e^{-y} = \log|x| - \frac{x^2}{2} + C \]Q5
\( e^x \sqrt{1-y^2} dx + \frac{y}{x} dy = 0 \)
Answer: \( e^x(x-1) = \sqrt{1-y^2} + C \)
Rearranging and separating variables:
\[ e^x \sqrt{1-y^2} dx = -\frac{y}{x} dy \] \[ x e^x dx = -\frac{y}{\sqrt{1-y^2}} dy \]Integrating both sides:
\[ \int x e^x dx = \int \frac{-y}{\sqrt{1-y^2}} dy \]LHS: Using integration by parts (\( u=x, dv=e^x dx \)):
\[ x e^x - \int e^x dx = x e^x - e^x = e^x(x-1) \]RHS: Let \( 1-y^2 = t \Rightarrow -2y dy = dt \Rightarrow -y dy = dt/2 \).
\[ \int \frac{1}{2\sqrt{t}} dt = \sqrt{t} = \sqrt{1-y^2} \]Combining results:
\[ e^x(x-1) = \sqrt{1-y^2} + C \]Q6
Find the equation of the curve passing through the point \( (1, -1) \) whose differential equation is \( xy \frac{dy}{dx} = (x+2)(y+2) \).
Answer: \( y - x + 2 = \log|x^2(y+2)^2| \)
Separating variables:
\[ \frac{y}{y+2} dy = \frac{x+2}{x} dx \] \[ \left( 1 - \frac{2}{y+2} \right) dy = \left( 1 + \frac{2}{x} \right) dx \]Integrating:
\[ y - 2\log|y+2| = x + 2\log|x| + C \]Using point \( (1, -1) \):
\[ -1 - 2\log|-1+2| = 1 + 2\log|1| + C \] \[ -1 - 2(0) = 1 + 0 + C \Rightarrow C = -2 \]Substituting \( C \):
\[ y - 2\log|y+2| = x + 2\log|x| - 2 \] \[ y - x + 2 = 2\log|x| + 2\log|y+2| \] \[ y - x + 2 = \log|x^2(y+2)^2| \]Q7
Solve \( (x+1)\frac{dy}{dx} = 2xy \), given that \( y(2) = 3 \).
Answer: \( \log|y| = 2x - 2\log|x+1| + 3\log 3 - 4 \)
Separating variables:
\[ \frac{dy}{y} = \frac{2x}{x+1} dx = \frac{2(x+1)-2}{x+1} dx \] \[ \frac{dy}{y} = \left( 2 - \frac{2}{x+1} \right) dx \]Integrating:
\[ \log|y| = 2x - 2\log|x+1| + C \]Using \( y(2)=3 \):
\[ \log 3 = 2(2) - 2\log(3) + C \] \[ \log 3 = 4 - 2\log 3 + C \Rightarrow C = 3\log 3 - 4 \]Solution:
\[ \log|y| = 2x - 2\log|x+1| + 3\log 3 - 4 \]Q8
Find the particular solution of the differential equation \( \log(\frac{dy}{dx}) = 3x + 4y \), given that \( y=0 \), when \( x=0 \).
Answer: \( 4e^{3x} + 3e^{-4y} - 7 = 0 \)
Rewrite equation:
\[ \frac{dy}{dx} = e^{3x+4y} = e^{3x} \cdot e^{4y} \]Separating variables:
\[ \frac{dy}{e^{4y}} = e^{3x} dx \Rightarrow e^{-4y} dy = e^{3x} dx \]Integrating:
\[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C \]Using \( x=0, y=0 \):
\[ \frac{e^0}{-4} = \frac{e^0}{3} + C \] \[ -\frac{1}{4} = \frac{1}{3} + C \Rightarrow C = -\frac{1}{4} - \frac{1}{3} = -\frac{7}{12} \]Substitute \( C \):
\[ -\frac{e^{-4y}}{4} = \frac{e^{3x}}{3} - \frac{7}{12} \]Multiply by -12:
\[ 3e^{-4y} = -4e^{3x} + 7 \] \[ 4e^{3x} + 3e^{-4y} - 7 = 0 \]