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daser333 [38]
3 years ago
7

Can someone help me solving this differential equation?

Mathematics
1 answer:
maria [59]3 years ago
8 0

Use reduction of order. Given a solution y_1(x) = x^2, look for a second solution of the form y_2(x) = y_1(x)v(x).

Compute the first two derivatives of y_2(x):

y_2 = x^2v \\\\ {y_2}' = x^2v' + 2xv \\\\ {y_2}'' = x^2v''+4xv' + 2v

Substitute them into the ODE:

x^4 (x^2v'' + 4xv' + 2v) + x^3 (x^2v' + 2xv) - 4x^2 (x^2v) = 1 \\\\ x^6v'' + 5x^5v' = 1

Now substitute w(x) = v'(x) and you end up with a linear ODE:

x^6w'+5x^5w=1

Multiply through both sides by \frac1x (if you're familiar with the integrating factor method, this is it):

x^5w'+5x^4w = \dfrac1x

Bear in mind that in order to do this, we require x\neq0. Just to avoid having to deal with absolute values later, let's further assume x>0.

Notice that the left side is the derivative of a product,

\left(x^5w\right)' = \dfrac1x

Integrate both sides with respect to x :

x^5w = \displaystyle \int\frac{\mathrm dx}x \\\\ x^5w = \ln(x) + C_1

Solve for w(x) :

w = \dfrac{\ln(x)+C_1}{x^5}

Solve for v(x) by integrating both sides:

v = \displaystyle \int \frac{\ln(x)+C_1}{x^5} \,\mathrm dx

Integrate by parts:

\displaystyle f = \ln(x) + C_1 \implies \mathrm df = \frac{\mathrm dx}x \\\\ \mathrm dg = \frac{\mathrm dx}{x^5} \implies g = -\frac1{4x^4} \\\\ \implies v = -\frac{\ln(x)+C_1}{4x^4} + \frac14 \int \frac{\mathrm dx}{x^5} \\\\ v = -\frac{\ln(x)+C_1}{4x^4} - \frac1{16x^4} + C_2 \\\\ v = -\frac{4\ln(x)+C_1}{16x^4}+C_2

Solve for y_2(x) :

\displaystyle \frac{y_2}{x^2} = -\frac{4\ln(x)+C_1}{16x^4}+C_2 \\\\ y_2 = -\frac{4\ln(x)+C_1}{16x^2} + C_2x^2

But since y_1(x)=x^2 is already accounted for, the second solution is just

\displaystyle y_2 = -\frac{4\ln(x)+C_1}{16x^2}

Still, the general solution would be

\displaystyle \boxed{y(x) = -\frac{4\ln(x)+C_1}{16x^2} + C_2x^2}

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