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Identify the order of the differential equation.

Mathematics

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2 parts
a(iv)

Order 2, degree 2\text{Order 2, degree 2}

b(iv)

Homogeneous\text{Homogeneous}

Step-by-step solution

5 steps

Part a· 3 steps

The given differential equation is (y′′)2+xy′+…(y'')^2 + xy' + \dots.

  1. Identify the order of the differential equation.

    The order of a differential equation is the order of the highest derivative present in the equation. In this equation, the highest derivative is y′′y'', which is a second-order derivative.
    So, the order is 2.

  2. Identify the degree of the differential equation.

    The degree of a differential equation is the power of the highest order derivative, provided the equation is a polynomial in derivatives. In this equation, the highest order derivative is y′′y'', and its power is 2.
    So, the degree is 2.

  3. Select the correct option.

    The differential equation has Order 2 and Degree 2.
    Comparing with the given options:

Part a(i)

Order 1, degree 2

Part a(ii)

Order 2, degree 2

Part a(iii)

Order 2, degree 1

Part a(iv)

Order 2, degree 2\text{Order 2, degree 2}

Order 1, degree 1

The correct option is (ii).

The final answer is Order 2, degree 2\boxed{\text{Order 2, degree 2}}

Part b· 2 steps

The given differential equation is y′=x+yx−yy' = \frac{x+y}{x-y}.

  1. Check if the equation is homogeneous.

    A first-order differential equation dydx=f(x,y)\frac{dy}{dx} = f(x,y) is homogeneous if f(tx,ty)=f(x,y)f(tx, ty) = f(x,y) for any non-zero constant tt.
    Let f(x,y)=x+yx−yf(x,y) = \frac{x+y}{x-y}.
    Substitute txtx for xx and tyty for yy:

    f(tx,ty)=tx+tytx−ty=t(x+y)t(x−y)=x+yx−yf(tx, ty) = \frac{tx+ty}{tx-ty} = \frac{t(x+y)}{t(x-y)} = \frac{x+y}{x-y}

    Since f(tx,ty)=f(x,y)f(tx, ty) = f(x,y), the differential equation is homogeneous.

  2. Select the correct option.

    Comparing with the given options:

Part b(i)

Exact

Part b(ii)

Homogeneous

Part b(iii)

Linear

Part b(iv)

Homogeneous\text{Homogeneous}

Bernoulli

The equation is homogeneous.

The final answer is Homogeneous\boxed{\text{Homogeneous}}

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