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Match each differential equation to a function which is a solution

FUNCTIONS
A. y = 3x + x^2,
B. y = e^{-3 x},
C. y = \sin(x),
D. y = x^{\,\frac{1}{2}},
E. y = 5 \exp(5 x),

DIFFERENTIAL EQUATIONS

1. y'' + 8 y' + 15 y = 0
2. y'' + y = 0
3. y' = 5 y
4. 2x^2y'' + 3xy' = y
Best Answer

Respuesta :

Kazeemsodikisola

Answer:

1. First equation is option B

2. Second equation is option C

3. Third equation is option E

4. Fourth equation no best option.

explanation:

Check the attachment for solution

Ver imagen Kazeemsodikisola
Ver imagen Kazeemsodikisola

Following are the calculation to the differential equation:

For point 1)

[tex]y'' + 8 y' + 15 y = 0\\[/tex]

B

[tex]Y = e^{-3x}[/tex] be the solution of this equation

[tex]Y' = -3 e^{-3x}\\\\ y''= 9 e^{-3x} \\\\\therefore \\\\y'' +8y' + 15 y= 9e^{-3x} + 8(-3e^{-3x})+ 15 e^{-3x} \\\\e^{-3x}( 9-24+15)=0[/tex]

For point 2)  

[tex]y'' + y = 0[/tex]

C

[tex]y = \sin x[/tex] be the solution of above equation  

[tex]y'= -\cos x \\\\y''= -\sin x = -y \\\\y''+y=0\\\\[/tex]

For point 3)

 [tex]y' = 5 y[/tex]

[tex]y'=e^{5x}[/tex] be the solution of equation 3

[tex]y'= 5 e^{5y}= 5y =y'=5y[/tex]

For point 4)

[tex]2x^2 y'' + 3xy' = y[/tex]

Let [tex]y=\sqrt{x}[/tex] be the solution of equation  (4)

 [tex]y'=\frac{1}{2 \sqrt{x} }\\\\y''=- \frac{1}{2} \times \frac{1}{2} \times {x^{- \frac{3}{2}}} ==- \frac{1}{4} \times {x^{- \frac{3}{2}}} \\\\-2x^2 \times =- \frac{1}{4} {x^{- \frac{3}{2}}}+ 3x \times =- \frac{1}{2 \sqrt{x}}\\\\- \frac{1}{2} {x^{ \frac{1}{2}}}+ \frac{3}{2} x^{\frac{1}{2}} =\sqrt{x} =y\\\\[/tex]

Learn more:

brainly.com/question/20479450

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