Question

Помогите, пожалуйста, нужно решить дифференциальные уравнения первого и второго порядка:

Answer 1

Ответ:

$y=e^{3x}(C_{1}\cos x+C_{2}\sin x), \ C_{1}, C_{2}-const;$

$y=1\dfrac{4}{7}e^{-4x}+1\dfrac{3}{7}e^{3x};$

Пошаговое объяснение:

$4) \quad y''-6y'+10y=0;$

$\lambda^{2}-6\lambda+10=0;$

$D=b^{2}-4ac \Rightarrow D=(-6)^{2}-4 \cdot 1 \cdot 10=36-40=-4;$

$\lambda_{1,2}=\dfrac{-b \pm \sqrt{D}}{2a} \Rightarrow \lambda_{1,2}=\dfrac{-(-6) \pm \sqrt{-4}}{2 \cdot 1}=\dfrac{6 \pm 2i}{2}=3 \pm i;$

$\lambda_{1,2}=3 \pm i \Rightarrow \alpha=3, \ \beta=1;$

$y=e^{\alpha x}(C_{1}\cos\beta x+C_{2}\sin\beta x);$

$y=e^{3x}(C_{1}\cos x+C_{2}\sin x), \ C_{1}, C_{2}-const;$

$5) \quad y''+y'-12y=0;$

$\lambda^{2}+\lambda-12=0;$

$\displaystyle \left \{ {{\lambda_{1}+\lambda_{2}=-1} \atop {\lambda_{1} \cdot \lambda_{2}=-12}} \right. \Leftrightarrow \left \{ {{\lambda_{1}=-4} \atop {\lambda_{2}=3}} \right. ;$

$y=C_{1}e^{\lambda_{1}x}+C_{2}e^{\lambda_{2}x};$

$y=C_{1}e^{-4x}+C_{2}e^{3x}, \ C_{1}, C_{2}-const;$

$y'=-4C_{1}e^{-4x}+3C_{2}e^{3x};$

$y=3, \quad y'=-2, \quad x=0;$

$\displaystyle \left \{ {{C_{1}e^{-4 \cdot 0}+C_{2}e^{3 \cdot 0}=3} \atop {-4C_{1}e^{-4 \cdot 0}+3C_{2}e^{3 \cdot 0}=-2}} \right. \Leftrightarrow \left \{ {{C_{1}+C_{2}=3} \atop {-4C_{1}+3C_{2}=-2}} \right. \Leftrightarrow \left \{ {{4C_{1}+4C_{2}=12} \atop {-4C_{1}+3C_{2}=-2}} \right. \bigg |+$

$7C_{2}=10 \Rightarrow C_{2}=\dfrac{10}{7}=1\dfrac{3}{7}; \quad C_{1}=3-1\dfrac{3}{7}=1\dfrac{4}{7};$

$y=1\dfrac{4}{7}e^{-4x}+1\dfrac{3}{7}e^{3x};$