Unfamiliar Equation

Calculus Level 5

2 d 2 y d x 2 + 3 d y d x + y = e − 3 x , x ∈ R 2 \dfrac{d^2y}{dx^2} + 3 \dfrac{dy}{dx} + y = e^{-3x} , \quad x \in \mathbb{R}

Let y ⁣ : R → R y \colon \mathbb{R} \to \mathbb{R} be a solution of the above ordinary differential equation, satisfying lim ⁡ x → ∞ e x y ( x ) = 0 \displaystyle\lim_{x\rightarrow \infty} e^xy(x) = 0 . Find y ( 0 ) y(0) .


The answer is 0.1.

Ravneet Singh by Ravneet Singh , 30, India

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2 solutions

Chew-Seong Cheong
Apr 21, 2017

The characteristic equation of the ordinary differential equation is:

2 D 2 + 3 D + 1 = 0 ( 2 D + 1 ) ( D + 1 ) = 0 ⟹ D = − 1 2 , − 1 \begin{aligned} 2D^2+3D+1&=0 \\ (2D+1)(D+1)&=0 \\ \implies D &= -\frac 12, \ -1 \end{aligned}

Therefore, the general solution of the ODE is y ( x ) = c 1 e − x 2 + c 2 e − x y(x)=c_1e^{-\frac x2}+c_2e^{-x} .

Assuming the particular solution of the ODE be:

y ( x ) = c e − 3 x y ′ ( x ) = − 3 c e − 3 x y ′ ′ ( x ) = 9 c e − 3 x ⟹ 2 y ′ ′ ( x ) + 3 y ′ ( x ) + 1 = 10 c e − 3 x 2 y ′ ′ ( 0 ) + 3 y ′ ( 0 ) + 1 = 10 c = 1 ⟹ c = 1 10 \begin{aligned} y(x) &=ce^{-3x} \\ y'(x)&= -3ce^{-3x} \\ y''(x)&= 9ce^{-3x} \\ \implies 2y''(x)+3y'(x)+1&= 10ce^{-3x} \\ 2y''(0)+3y'(0)+1&= 10c = 1 \\ \implies c &=\frac 1{10} \end{aligned}

Therefore, the particular solution of the ODE is y ( x ) = e − 3 x 10 y(x)=\frac {e^{-3x}}{10} .

And the complete solution of the ODE is:

y ( x ) = c 1 e − x 2 + c 2 e − x + e − 3 x 10 ⟹ e x y ( x ) = c 1 e x 2 + c 2 + e − 2 x 10 Since lim ⁡ x → ∞ e x y ( x ) = 0 ⟹ c 1 = c 2 = 0 ⟹ y ( x ) = e − 3 x 10 ⟹ y ( 0 ) = 1 10 = 0.1 \begin{aligned} y(x)&=c_1e^{-\frac x2}+c_2e^{-x} + \frac {e^{-3x}}{10} \\ \implies e^xy(x) &= c_1e^{\frac x2}+c_2 + \frac {e^{-2x}}{10} & \small \color{#3D99F6} \text{Since }\lim_{x \to \infty} e^xy(x)=0 \\ \implies c_1 &=c_2 =0 \\ \implies y(x)&=\frac {e^{-3x}}{10} \\ \implies y(0)&=\frac 1{10} =\boxed{0.1} \end{aligned}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

Well explained.... Thanks (+1)

Ravneet Singh - 4 years, 1 month ago
Guilherme Niedu
Apr 21, 2017

One possible solution is:

y ( x ) = c e − 3 x \large \displaystyle y(x) = ce^{-3x}

Plugging this into the equation:

18 c e − 3 x − 9 c e − 3 x + c e − 3 x = e − 3 x \large \displaystyle 18ce^{-3x} - 9ce^{-3x} + ce^{-3x} = e^{-3x}

10 c e − 3 x = e − 3 x \large \displaystyle 10ce^{-3x} = e^{-3x}

10 c = 1 \large \displaystyle 10c = 1

c = 0.1 \large \displaystyle c = 0.1

y ( x ) = 0.1 e − 3 x \color{#20A900} \boxed{\large \displaystyle y(x) = 0.1e^{-3x}}

So:

y ( 0 ) = 0.1 \color{#3D99F6} \boxed{\large \displaystyle y(0) = 0.1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, the particular solution y p = c e − 3 x \large y_{p} = ce^{-3x} ends up being all we need to consider. The homogenous solution is y h = a e − x / 2 + b e − x \large y_{h} = ae^{-x/2} + be^{-x} , and so the general solution is

y = y h + y p = a e − x / 2 + b e − x + c e − 3 x \large y = y_{h} + y_{p} = ae^{-x/2} + be^{-x} + ce^{-3x} , and so

lim ⁡ x → ∞ e x y = lim ⁡ x → ∞ ( a e x / 2 + b + c e − 2 x ) = 0 \large \displaystyle \lim_{x \to \infty} e^{x}y = \lim_{x \to \infty} (ae^{x/2} + b + ce^{-2x}) = 0 only if a = b = 0 a = b = 0 .

Brian Charlesworth - 4 years, 1 month ago

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