7BSolutionsOfDEs


 * Differential Equations**

A **Differential Equation (D.E.)** is an equation that has at least one derivative.

Example math . \qquad \dfrac{dy}{dx}+3y=2x math

The **Order** of a D.E. is given by the highest degree of derivative in the equation.

Examples math \\ . \qquad 3x\dfrac{dy}{dx}=2 \quad \text{ is a } \textbf{1st Order } \text{differential equation} \\ .\\ . \qquad \dfrac{d^2y}{dx^2}+3\dfrac{dy}{dx}+2y=0 \quad \text{ is a } \textbf{2nd Order } \text{differential equation} math

Verifying Solutions of Differential Equations

To verify that a function, y = f(x) is a solution to a D.E. the function must be substituted into the left side and simplifed to gain the right side.

Example 1

math \text{Show that } \, y=\big( 2x+3 \big) e^{-3x} \, \text{ is a solution to } \, \dfrac{d^2y}{dx^2}+6\dfrac{dy}{dx}+9y=0 math

Find the first derivative math . \qquad y=\big( 2x+3 \big) e^{-3x} math . math . \qquad \dfrac{dy}{dx} = 2e^{-3x}-3 \big( 2x+3 \big) e^{-3x} \quad \textit{ using product rule} math . math . \qquad \quad = \big( 2-6x-9 \big) e^{-3x} \qquad \textit{taking out } e^{-3x} \textit{ as a common factor} math . math . \qquad \quad = \big( -6x-7 \big) e^{-3x} math

Now find the second derivative math . \qquad \dfrac{d^2y}{dx^2} = -6e^{-3x}-3 \big( -6x-7 \big) e^{-3x} \quad \textit{ using product rule} math . math . \qquad \quad \; = \big( -6+18x+21 \big) e^{-3x} \qquad \textit{taking out } e^{-3x} \textit{ as a common factor} math . math . \qquad \quad \; = \big( 18x+15 \big) e^{-3x} math

Now to verify the equation math . \qquad LHS = \dfrac{d^2y}{dx^2}+6\dfrac{dy}{dx}+9y math . math . \qquad \qquad = \big( 18x+15 \big) e^{-3x} + 6 \big( -6x-7 \big) e^{-3x} + 9 \big( 2x+3 \big) e^{-3x} math . math .\qquad \qquad = \big( 18x+15-36x-42+18x+27 \big) e^{-3x} math . math \\ .\qquad \qquad = \big( 0 \big) e^{-3x} \\, \\ . \qquad \qquad = 0 \\. \\ . \qquad \qquad = RHS math

math \text{Hence} \, y=\big( 2x+3 \big) e^{-3x} \, \text{ is a solution to } \, \dfrac{d^2y}{dx^2}+6\dfrac{dy}{dx}+9y=0 \, \text{ as required.} math

Example 2

math \\ \text{Find the values of } \, m \, \text{ for which } \, y=2e^{mx} \\. \\ \text{is a solution to } \, \dfrac{d^2y}{dx^2}+9\dfrac{dy}{dx}+20y=0 math

First, find the first and second derivatives math \\ . \qquad y=2e^{mx} \\. \\ . \qquad \dfrac{dy}{dx} = 2me^{mx} \\. \\ . \qquad \dfrac{d^2y}{dx^2} = 2m^2e^{mx} math

Substitute these into the D.E. and solve for m math . \qquad \dfrac{d^2y}{dx^2}+9\dfrac{dy}{dx}+20y=0 math . math . \qquad 2m^2e^{mx}+9 \times 2me^{mx}+20 \times 2e^{mx} = 0 math . math . \qquad 2m^2e^{mx}+18me^{mx}+40e^{mx} = 0 math . math . \qquad 2e^{mx} \big( m^2+9m+20 \big) =0 math . math . \qquad m^2+9m+20=0 \; \textbf{ or } \; 2e^{mx}=0 \quad \textit{ The second option can't happen, so discard it} math , math , \qquad (m+5)(m+4)=0 math . math . \qquad m=-5, \; m=-4 math

Typo in old text: Ex 7B Qn 8. DE: math . \qquad x \dfrac{d^2y}{dx^2} + 4x^3y = \dfrac{dy}{dx} math

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