Let's do some calculus! (56)

Calculus Level 5

Let f : R → R f: \mathbb{R} \to \mathbb{R} given by y = f ( x ) y = f(x) be a real-valued continuous function satisfying the differential equation

d 3 y d x 3 − d 2 y d x 2 − d y d x + y = 2 e x \large \dfrac{d^3 y}{dx^3} - \dfrac{d^2 y}{dx^2} - \dfrac{dy}{dx} + y = 2e^x

Given that f ( 0 ) = 0 , f ′ ( 0 ) = 1 f(0)=0, f'(0)=1 and f ′ ′ ( 0 ) = 1 f''(0)=1 . Find the solution of the differential equation.

For more problems on calculus, click here .
y = 1 2 ( x 2 e x + e x − e − x ) y= \dfrac12 \left(x^2 e^x + e^x - e^{-x}\right) y = 1 2 ( x e x + e x + e − x ) y= \dfrac12 \left(x e^x + e^x + e^{-x}\right) y = 1 2 ( x e x + e x − e − x ) y= \dfrac12 \left(x e^x + e^x - e^{-x}\right) y = 1 2 ( x 2 e x + e x + e − x ) y= \dfrac12 \left(x^2 e^x + e^x + e^{-x}\right)

Tapas Mazumdar by Tapas Mazumdar , 20, India

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3 solutions

Tom Engelsman
May 13, 2018

This can be accomplished via a Laplace Transform:

y ′ ′ ′ − y ′ ′ − y ′ + y = 2 e x ; y''' - y'' - y' + y = 2e^{x};

or L [ y ( x ) ] = ( s 3 Y ( s ) − s 2 y ( 0 ) − s y ′ ( 0 ) − y ′ ′ ( 0 ) ) − ( s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) ) − ( s Y ( s ) − y ( 0 ) ) + Y ( s ) = 2 s − 1 ; L[y(x)] = (s^3Y(s) - s^2 y(0) - s y'(0) - y''(0)) - (s^2 Y(s) - s y(0) - y'(0)) - (s Y(s) - y(0)) + Y(s) = \frac{2}{s-1};

or ( s 3 − s 2 − s + 1 ) Y ( s ) − s − 1 + 1 = 2 s − 1 ; (s^3 - s^2 - s + 1)Y(s) - s - 1 + 1 = \frac{2}{s-1};

or ( s 3 − s 2 − s + 1 ) Y ( s ) = 2 s − 1 + s ( s − 1 ) s − 1 ; (s^3 - s^2 - s + 1)Y(s) = \frac{2}{s-1} + \frac{s(s-1)}{s-1};

or Y ( s ) = s 2 − s + 2 ( s − 1 ) ( s 3 − s 2 − s + 1 ) = − 1 2 ( s + 1 ) + 1 2 ( s − 1 ) + 1 ( s − 1 ) 3 ; Y(s) = \frac{s^2 - s + 2}{(s-1)(s^3 - s^2 - s + 1)} = -\frac{1}{2(s+1)} + \frac{1}{2(s-1)} + \frac{1}{(s-1)^3};

or y ( x ) = L − 1 [ Y ( s ) ] = − 1 2 ⋅ e − x + 1 2 ⋅ e x + 1 2 ⋅ x 2 e x . y(x) = L^{-1}[Y(s)] = \boxed{-\frac{1}{2} \cdot e^{-x} + \frac{1}{2} \cdot e^{x} + \frac{1}{2} \cdot x^2 e^{x}}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Correction: Partial fraction decomposition of s 2 − s + 2 ( s − 1 ) ( s 3 − s 2 − s + 1 ) = − 1 2 ( s + 1 ) + 1 2 ( s − 1 ) + 1 ( s − 1 ) 3 \dfrac{s^2 - s + 2}{\left(s-1\right)\left(s^3 - s^2 - s + 1\right)} = -\dfrac1{2\left(s+1\right)} + \dfrac1{2\left(s-1\right)} +\dfrac1{\left(s-1\right)^3} meaning that you have an extra 2 2 in the last denominator. However, the inverse laplace transform of this term is correct in the final result.

Jesse Nieminen - 3 years ago

Nice work using Laplace Transformation! :)

Tapas Mazumdar - 3 years ago
Tapas Mazumdar
May 14, 2018

I'll state the differential coefficients as d n y d x n = f ′ ′ ′ ⋯ ′ ′ ′ (n times) ( x ) \dfrac{d^n y}{dx^n} = f^{''' \cdots ''' \ \text{ (n times) }} (x) .

f ′ ′ ′ ( x ) − f ′ ′ ( x ) − f ′ ( x ) + f ( x ) = 2 e x ⟹ e − x ( f ′ ′ ′ ( x ) − f ′ ′ ( x ) ) − e − x ( f ′ ( x ) − f ( x ) ) = 2 Integrate both sides using ∫ e x ( f ( x ) + f ′ ( x ) ) d x = e x f ( x ) + C ⟹ e − x f ′ ′ ( x ) − e − x f ( x ) = 2 x + c 1 Using f ′ ′ ( 0 ) = 1 and f ( 0 ) = 0 we get c 1 = 1 ⟹ e − x f ′ ′ ( x ) − e − x f ′ ( x ) + e − x f ′ ( x ) − e − x f ( x ) = 2 x + 1 ⟹ e − x ( f ′ ′ ( x ) − f ′ ( x ) ) + e − x ( f ′ ( x ) − f ( x ) ) = 2 x + 1 Integrate both sides ⟹ e − x f ′ ( x ) + e − x f ( x ) = x 2 + x + c 2 Using f ′ ( 0 ) = 1 and f ( 0 ) = 0 we get c 2 = 1 ⟹ e − x f ′ ( x ) + e − x f ( x ) = x 2 + x + 1 ⟹ f ′ ( x ) + f ( x ) = x 2 e x + x e x + e x Solving the simple Bernoulli LDE ⟹ f ( x ) e x = 1 2 ( x 2 e 2 x + e 2 x + c 3 ) Using f ( 0 ) = 0 we get c 3 = − 1 ⟹ y = 1 2 ( x 2 e x + e x − e − x ) \begin{aligned} & f'''(x) - f''(x) - f'(x) + f(x) = 2e^x \\ \implies & e^{-x} \left( f'''(x) - f''(x) \right) - e^{-x} \left( f'(x) - f(x) \right) = 2 & \small \color{#3D99F6}{\text{Integrate both sides using } \int e^x (f(x) + f'(x)) \,dx = e^x f(x) + C} \\ \implies & e^{-x} f''(x) - e^{-x} f(x) = 2x + c_1 & \small \color{#3D99F6}{\text{Using } f''(0) = 1 \text{ and } f(0) = 0 \text{ we get } c_1 = 1} \\ \implies & e^{-x} f''(x) - e^{-x} f'(x) + e^{-x} f'(x) - e^{-x} f(x) = 2x +1 \\ \implies & e^{-x} \left( f''(x) - f'(x) \right) + e^{-x} \left( f'(x) - f(x) \right) = 2x + 1 & \small \color{#3D99F6}{\text{Integrate both sides}} \\ \implies & e^{-x} f'(x) + e^{-x} f(x) = x^2 + x + c_2 & \small \color{#3D99F6}{\text{Using } f'(0) = 1 \text{ and } f(0) = 0 \text{ we get } c_2 = 1} \\ \implies & e^{-x} f'(x) + e^{-x} f(x) = x^2 + x + 1 \\ \implies & f'(x) + f(x) = x^2 e^x + x e^x + e^x & \small \color{#3D99F6}{\text{Solving the simple Bernoulli LDE}} \\ \implies & f(x) e^x = \dfrac{1}{2} \left( x^2 e^{2x} + e^{2x} + c_3 \right) & \small \color{#3D99F6}{\text{Using } f(0) = 0 \text{ we get } c_3 = -1} \\ \implies & \boxed{y = \dfrac{1}{2} \left( x^2 e^{x} + e^{x} - e^{-x} \right)} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Solving the Bernoulli LDE:

d y d x + y = x 2 e x + x e x + e x \dfrac{dy}{dx} + y = x^2 e^x + x e^x + e^x

Comparing this to d y d x + y P ( x ) = Q ( x ) \dfrac{dy}{dx} + y P(x) = Q(x) , we get P ( x ) = 1 P(x) =1 and Q ( x ) = x 2 e x + x e x + e x Q(x) = x^2 e^x + x e^x + e^x .

The integrating factor I ( x ) = e ∫ P ( x ) d x = e ∫ d x = e x \mathfrak{I} (x) = e^{ \int P(x) \,dx} = e^{ \int \,dx} = e^x .

The solution is thus

y I ( x ) = ∫ Q ( x ) ⋅ I ( x ) d x ⟹ y e x = ∫ ( x 2 e 2 x + x e 2 x + e 2 x ) d x ⟹ y e x = 1 2 ( x 2 e 2 x + e 2 x + C ) \begin{aligned} & y \mathfrak{I} (x) = \displaystyle \int Q(x) \cdot \mathfrak{I} (x) \,dx \\ \implies & y e^x = \displaystyle \int (x^2 e^{2x} + xe^{2x} + e^{2x}) \,dx \\ \implies & y e^x = \dfrac{1}{2} \left( x^2 e^{2x} + e^{2x} + C \right) \end{aligned}

Tapas Mazumdar - 3 years ago
Aaghaz Mahajan
May 13, 2018

@Tapas Mazumdar Kindly please upload the solution......or at least give a mere hint....

0 Helpful 0 Interesting 0 Brilliant 0 Confused

e x ( f ( x ) + f ′ ( x ) ) d x = e x f ( x ) + C \displaystyle e^x \left( f(x) + f'(x) \right) \, dx = e^x f(x) + C . There you go! :)

Tapas Mazumdar - 3 years ago

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