Question

Solve the following differential equation by reducing it to normal form using change of dependent variable
d^2y/dx^2-2xdy/dx + (x^2+2)y= e^1/2(x^2+2x)

Answer 1

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Solve ordinary differential equation :

$$\frac{d^2y}{dx^2}-2x\cdot\frac{dy}{dx} + (x^2+2)y= \sqrt{e}(x^2+2)$$

From the theory we know that for solving the equation

$$y''+f(x)\cdot y'+g(x)\cdot y=F(x)$$

convenient to introduce a new variable

$$y(x)=k(x)\cdot exp\left(-\frac{1}{2}\cdot\int f(x)dx\right)$$

In our case,

$$\frac{d^2y}{dx^2}\underbrace{-2x}_{f(x)}\cdot\frac{dy}{dx} +\underbrace{(x^2+2)}_{g(x)}y= \underbrace{\sqrt{e}(x^2+2)}_{F(x)}$$

Then,

$$f(x)=-2x\to-\frac{1}{2}\cdot\int f(x)dx=-\frac{1}{2}\cdot\int (-2x)dx=\frac{x^2}{2}\\[0.3cm] y(x)=k(x)\cdot e^{x^2/2}\to \frac{dy}{dx}=k'\cdot e^{x^2/2}+xk\cdot e^{x^2/2}\\[0.3cm] \boxed{\frac{dy}{dx}=e^{x^2/2}\cdot\left(k'+xk\right)}\\[0.3cm] \frac{d^2y}{dx^2}=k''\cdot e^{x^2/2}+xk'\cdot e^{x^2/2}+k\cdot e^{x^2/2}+xk'\cdot e^{x^2/2}+x^2k\cdot e^{x^2/2}\\[0.3cm] \boxed{\frac{d^2y}{dx^2}=e^{x^2/2}\cdot\left(k'' +2xk'+k+x^2k\right)}$$

We substitute the found derivatives into the initial equation

$$\frac{d^2y}{dx^2}-2x\cdot\frac{dy}{dx}+(x^2+2)\cdot y=\sqrt{e}\cdot(x^2+2)\\[0.3cm] e^{x^2/2}\cdot\left(k'' +2xk'+k+x^2k\right)-2x\cdot e^{x^2/2}\cdot\left(k'+xk\right)+\\[0.3cm] \left.+(x^2+2)\cdot e^{x^2/2}\cdot k=\sqrt{e}\cdot(x^2+2)\right|\div\left(e^{-x^2/2}\right)\\[0.3cm] k'' +2xk'+k+x^2k-2xk'-2x^2k+x^2k+2k=e^{-x^2/2}\cdot\sqrt{e}\cdot(x^2+2)\\[0.3cm] k''+3k=e^{-x^2/2}\cdot\sqrt{e}\cdot(x^2+2)$$

This is a second order inhomogeneous equation. The solutions will have the form :

$$k(x)=k_{hom}(x)+k_{part}(x),\,\,\,\text{where}\\[0.3cm] k_{hom}(x) : k''+3k=0\to k(x)=e^{\lambda x}\to\\[0.3cm] k'(x)=\lambda\cdot e^{\lambda x}\,\,\,\text{and}\,\,\, k''(x)=\lambda^2\cdot e^{\lambda x}\\[0.3cm] e^{\lambda x}\cdot(\lambda^2+3)=0\to \left[\begin{array}{l} \lambda_1=i\sqrt{3}\\[0.3cm] \lambda_2=-i\sqrt{3} \end{array}\right.\longrightarrow\\[0.3cm] k_{hom}(x)=A_1\cdot e^{i\sqrt{3}x}+A_2\cdot e^{-i\sqrt{3}x}\equiv C_1\cos\left(\sqrt{3}x\right)+C_2\sin\left(\sqrt{3}x\right)\\[0.3cm] \boxed{k_{hom}(x)= C_1\cos\left(\sqrt{3}x\right)+C_2\sin\left(\sqrt{3}x\right)}$$

We look for a particular solution in the form

$$k_{part}(x)=e^{-x^2/2}\cdot\left(Ax^2+Bx+C\right)\\[0.3cm] k''=\left(e^{-x^2/2}\right)''\cdot\left(Ax^2+Bx+C\right)+2\left(e^{-x^2/2}\right)'\left(Ax^2+Bx+C\right)'\\[0.3cm] +e^{-x^2/2}\left(Ax^2+Bx+C\right)'' =e^{-x^2/2}(x^2-1)\left(Ax^2+Bx+C\right)+\\[0.3cm]+e^{-x^2/2}(-2x)\left(2Ax+B\right) +e^{-x^2/2}(2A)\\[0.3cm] k''+3k=e^{-x^2/2}((x^2-1)\left(Ax^2+Bx+C\right)-4Ax^2-2Bx+2A)\to\\[0.3cm] e^{-x^2/2}((x^2-1)\left(Ax^2+Bx+C\right)-4Ax^2-2Bx+2A)+\\[0.3cm] \left.+3\cdot e^{-x^2/2}(Ax^2+Bx+C)=e^{(1-x^2)/2}(x^2+2)\right|\div\left(e^{-x^2/2}\right)\\[0.3cm] Ax^4+Bx^3+Cx^2-Ax^2-Bx-C-4Ax^2-2Bx+2A+\\[0.3cm] +3Ax^2+3Bx+3C=\sqrt{e}(x^2+2)\\[0.3cm] Ax^4+Bx^3+(C-2A-\sqrt{e})x^2+(2A+2C-2\sqrt{e})=0\\[0.3cm] \left\{\begin{array}{l} x^4 : A=0\\[0.3cm] x^3 : B=0\\[0.3cm] x^2 : C-2A-\sqrt{e}=0\to C=\sqrt{e}\\[0.3cm] x^0 : 2A+2C-2\sqrt{e}=0 \end{array}\right.$$

Conclusion,

$$\boxed{k_{part}(x)=e^{-x^2/2}\cdot\sqrt{e}}$$

Then,

$$k(x)=k_{hom}(x)+k_{part}(x)\to\\[0.3cm] \boxed{k(x)=C_1\cos\left(\sqrt{3}x\right)+C_2\sin\left(\sqrt{3}x\right)+e^{-x^2/2}\cdot\sqrt{e}}$$

Finally, general solution is

$$y(x)=e^{x^2/2}\cdot k(x)=\\[0.3cm] =\left(C_1\cos\left(\sqrt{3}x\right)+C_2\sin\left(\sqrt{3}x\right)+e^{-x^2/2}\cdot\sqrt{e}\right)\cdot e^{x^2/2}\to\\[0.3cm] \boxed{y(x)=\left(C_1\cos\left(\sqrt{3}x\right)+C_2\sin\left(\sqrt{3}x\right)\right)\cdot e^{x^2/2}+\sqrt{e}}$$

ANSWER

$$y(x)=\left(C_1\cos\left(\sqrt{3}x\right)+C_2\sin\left(\sqrt{3}x\right)\right)\cdot e^{x^2/2}+\sqrt{e}$$