a) The solution of this ordinary differential equation is [tex]y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }[/tex].
b) The integrating factor for the ordinary differential equation is [tex]-\frac{1}{x}[/tex].
The particular solution of the ordinary differential equation is [tex]y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}[/tex].
How to solve ordinary differential equations
a) In this case we need to separate each variable ([tex]y, t[/tex]) in each side of the identity:
[tex]6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t[/tex] (1)
[tex]6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C[/tex]
Where [tex]C[/tex] is the integration constant.
By table of integrals we find the solution for each integral:
[tex]-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C[/tex]
If we know that [tex]x = 0[/tex] and [tex]y = 1[/tex], then the integration constant is [tex]C = -2[/tex].
The solution of this ordinary differential equation is [tex]y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }[/tex]. [tex]\blacksquare[/tex]
b) In this case we need to solve a first order ordinary differential equation of the following form:
[tex]\frac{dy}{dx} + p(x) \cdot y = q(x)[/tex] (2)
Where:
- [tex]p(x)[/tex] - Integrating factor
- [tex]q(x)[/tex] - Particular function
Hence, the ordinary differential equation is equivalent to this form:
[tex]\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}[/tex] (3)
The integrating factor for the ordinary differential equation is [tex]-\frac{1}{x}[/tex]. [tex]\blacksquare[/tex]
The solution for (2) is presented below:
[tex]y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C[/tex] (4)
Where [tex]C[/tex] is the integration constant.
If we know that [tex]p(x) = -\frac{1}{x}[/tex] and [tex]q(x) = x^{2} + \frac{1}{x}[/tex], then the solution of the ordinary differential equation is:
[tex]y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C[/tex]
[tex]y = x\int {x} \, dx + x\int\, dx + C[/tex]
[tex]y = \frac{x^{3}}{2}+x^{2}+C[/tex]
If we know that [tex]x = 1[/tex] and [tex]y = -1[/tex], then the particular solution is:
[tex]y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}[/tex]
The particular solution of the ordinary differential equation is [tex]y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}[/tex]. [tex]\blacksquare[/tex]
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