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Aleksandr-060686 [28]
2 years ago
11

Solve the following differential equation: y" + y' = 8x^2

Mathematics
1 answer:
PtichkaEL [24]2 years ago
5 0

Answer:

y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}

Step-by-step explanation:

Let's  find a particular solution. We need a function of the form y= ax^3+bx^2+cx+d such that

y'= 3ax^2+2bx+c and

y''= 6ax+2b

y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a =  8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is  y_p= \frac{8}{3}x^3-8x^2+16x.

Now, let's find the solution y_p of the homogeneus equation y''+y'=0 with the method of constants coefficients. Let y=e^{\lambda x}

y'=\lambda e^{\lambda x}

y''=\lambda^2e^{\lambda x}

then \lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0

e^{\lambda x}(\lambda +\lambda^2)= 0

(\lambda +\lambda^2)= 0

\lambda (1+\lambda)= 0

\lambda =0 and \lambda)= -1.

So, y_h = C_1 + C_2e^{-x} and the solution is

y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}.

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