Question

Apply the method of variations of parameters to solve the following differential equations:
a) x^2y''+xy'-y=x^2e^x
b)y"+a^2y=cosecax

Answer 1

a). At first, we solve the following differential equation:

$x^2y''+xy'-y=0$

We look for solution in the following form:

$y=x^{\alpha}.$

After substitution we receive:

$\alpha(\alpha-1)x^{\alpha}+\alpha x^{\alpha}-x^{\alpha}=0$

From this we receive:

$\alpha^2-1=0.$

Thus, we have the solution $y=c_1x+c_2\frac{1}{x}$ , $c_1,c_2\in\mathbb{C}$ .

Now, we set $y=c_1(x)x+c_2(x)\frac{1}{x}$ and add additional restriction:

$c_1'x+c_2'\frac1x=0$. The latter implies $c_1''x+c_1'+c_2''\frac1{x}-c_2'\frac1{x^2}=0$ . We substitute $y$ into $x^2y''+xy'-y=x^2e^x$

and get:

$x^2(c_1''x+2c_1'+c_2''\frac1x-2c_2'\frac1{x^2}+2c_2\frac1{x^3})+x(c_1'x+c_1+c_2'\frac1{x}-c_2\frac{1}{x^2})-(c_1x+c_2\frac{1}{x})=x^2e^x$

$x^2(c_1'-c_2'\frac1{x^2})=x^2e^x$

Thus, we receive a system of equations:

$x^2(c_1'-c_2'\frac1{x^2})=x^2e^x$

$c_1'x+c_2'\frac1x=0$

From the latter we obtain:

$2c_1'x=xe^x$

Thus, $c_1(x)=\frac12e^x+k_1$ , $c_2(x)=-\frac12(2-2x+x^2)e^x+k_2$ , $k_1,k_2\in\mathbb{C}$

The resulting solution is $y=(\frac12e^x+k_1)x+(-\frac12(2-2x+x^2)e^x+k_2)\frac{1}{x}$ .

b). At first, we solve the equation: $y''+a^2y=0$ . The solution is $y=c_1e^{i a x}+c_2e^{-i a x}$

We suppose that $y=c_1(x)e^{i a x}+c_2(x)e^{-i a x}$ and $c_1'e^{i a x}+c_2'e^{-i a x}=0$ . From the latter we get: $c_1''e^{i a x}+iac_1'e^{iax}+c_2''e^{-i a x}-iac_2'e^{-iax}=0$ We substitute $y$ into $y''+a^2y=\frac{1}{sin \,ax}$ and receive $c_1''e^{i a x}+2i\,a\,c_1'e^{i a x}-c_1a^2e^{i a x}+c_2''e^{-i a x}-2iac_2'e^{-i a x}-c_2a^2e^{-i a x}+a^2(c_1(x)e^{i a x}+c_2(x)e^{-i a x})=\frac{1}{sin \,ax}$

$i\,a\,c_1'e^{i a x}-iac_2'e^{-i a x}=\frac{1}{sin \,ax}$

$c_1'e^{i a x}+c_2'e^{-i a x}=0$

$2i\,a\,c_1'e^{i a x}=\frac{1}{sin \,ax}$

$c_1'=-i\frac{e^{-i a x}}{2a\,sin \,ax}$

$c_1=-\frac{i}{2a}(\frac{1}{a}ln(sin\,ax)-ix+c)$

$c_2=\frac{i}{2a}(\frac{1}{a}ln(sin\,ax)+ix+c)$

$y=-\frac{i}{2a}(\frac{1}{a}ln(sin\,ax)-ix+c)e^{i a x}+\frac{i}{2a}(\frac{1}{a}ln(sin\,ax)+ix+c)e^{-i a x}$

Answer: a) $y=(\frac12e^x+k_1)x+(-\frac12(2-2x+x^2)e^x+k_2)\frac{1}{x}$

b) $y=-\frac{i}{2a}(\frac{1}{a}ln(sin\,ax)-ix+c)e^{i a x}+\frac{i}{2a}(\frac{1}{a}ln(sin\,ax)+ix+c)e^{-i a x}$