Question

6 The solution to the differential equation
(4+t2)dydt+2ty=4t
is
y=4t24+t2+c6+t2
y=2t34+t3+c4+t2
y=2t27+t2
y=2t24+t2+c4+t2
7 Find the general solution of the differential equation
dydx−2y=4−x
y=−12+23x+ce2x
y=−74+12x+ce2x
y=−74+22x+ce3x
y=−ln74+12x+celnx
8 Solve the initial value problem
xdydx+2y=4x2
,
y(1)=2
y=x2+1x2,x>0
y=3x2+1x2,x>0
y=x−2+1x2,x>0
y=x3+13x2,x>0

9 Solve for
dydx+y=5sin2x
y=ce2x−sin2x−2cos3x
y=ce−x+tan2x−2cos2x
y=ce−x+sin2x+3cos2x
y=ce−x+sin2x−2cos2x

10 Given the differential equation
dydx+(1x)y=3cos2x,x>0
the solution is __
y=cx+3cos2x4x+3sin22
y=cx+3cos2x4x
y=cx+3sin22
y=c2x+3cosx4x+3sin62

Answer 1

Answer on Question #58420 – Math – Differential Equations

Question

6. The solution to the differential equation $(4 + t^{2})\frac{dy}{dt} + 2ty = 4t$ is

$$y = 4t24 + t2 + c6 + t2$$ $$y = 2t34 + t3 + c4 + t2$$ $$y = 2t27 + t2$$ $$y = 2t24 + t2 + c4 + t2$$

Solution

$$(4 + t^{2})\frac{dy}{dt} + 2ty = 4t$$ $$(4 + t^{2})dy = 2t(2 - y)dt$$ $$\frac{dy}{2 - y} = \frac{2t}{4 + t^{2}} dt$$ $$\int \frac{dy}{2 - y} = \int \frac{2t}{4 + t^{2}} dt$$ $$- \ln(2 - y) = \ln(t^{2} + 4) + c_{1}$$ $$\frac{c_{2}}{2 - y} = t^{2} + 4$$ $$\frac{c_{2}}{t^{2} + 4} = 2 - y$$ $$y = 2 - \frac{c_{2}}{t^{2} + 4} = \frac{2t^{2} + 8 - c_{2}}{t^{2} + 4} = \frac{2t^{2}}{t^{2} + 4} + \frac{8 - c_{2}}{t^{2} + 4}$$

Answer: $y = \frac{2t^{2}}{t^{2} + 4} + \frac{c}{t^{2} + 4}$

Question

7. Find the general solution of differential equation $\frac{dy}{dx} - 2y = 4 - x$

$$y = -12 + 23x + ce2x$$ $$y = -74 + 12x + ce2x$$ $$y = -74 + 22x + ce3x$$ $$y = -\ln(74 + 12x) + ce\ln x$$

Solution

$$\begin{array}{l} \frac{dy}{dx} - 2y = 4 - x \\ y = uv; \quad y' = u'v + uv' \end{array}$$ $$u'v + uv' - 2uv = 4 - x$$ $$u'v + u(v' - 2v) = 4 - x$$ $$\left\{ \begin{array}{l} v' - 2v = 0 \\ u'v = 4 - x \end{array} \right.$$ $$v' - 2v = 0 \Rightarrow \frac{dv}{dx} = 2v \quad \Rightarrow \quad \int \frac{dv}{2v} = \int dx \quad \Rightarrow \quad \frac{\ln v}{2} = x \quad \Rightarrow \quad v = e^{2x}$$ $$u' \cdot e^{2x} = 4 - x$$ $$u = \int \frac{4 - x}{e^{2x}} dx = \frac{2x - 7}{4e^{2x}} + c$$ $$y = uv = e^{2x} \left( \frac{2x - 7}{4e^{2x}} + c \right)$$

Answer: $y = ce^{2x} + \frac{x}{2} - \frac{7}{4}$.

Question

8. Solve the initial value problem $x \frac{dy}{dx} + 2y = 4x^2, \ y(1) = 2$.

$$\begin{array}{l} y = x^2 + 1x^2, x > 0 \\ y = 3x^2 + 1x^2, x > 0 \\ y = x - 2 + 1x^2, x > 0 \\ y = x^3 + 13x^2, x > 0 \end{array}$$

Solution

$$x \frac{dy}{dx} + 2y = 4x^2$$ $$y' + \frac{2}{x}y = 4x$$ $$y = u v ; \quad y ^ {\prime} = u ^ {\prime} v + u v ^ {\prime}$$ $$u ^ {\prime} v + u v ^ {\prime} + \frac {2 u v}{x} = 4 x$$ $$u ^ {\prime} v + u \left(v ^ {\prime} + \frac {2 v}{x}\right) = 4 x$$ $$\left\{ \begin{array}{l} v ^ {\prime} + \frac {2 v}{x} = 0 \\ u ^ {\prime} v = 4 x \end{array} \right.$$ $$\frac {d v}{d x} = - \frac {2 v}{x} \quad \Rightarrow \quad \int \frac {d v}{2 v} = - \int \frac {d x}{x} \quad \Rightarrow \quad \frac {\ln v}{2} = - \ln x \quad \Rightarrow \quad v = \frac {1}{x ^ {2}}$$ $$u ^ {\prime} \cdot \frac {1}{x ^ {2}} = 4 x$$ $$u = \int 4 x ^ {3} d x = x ^ {4} + c$$ $$y = u v = \left(x ^ {4} + c\right) \cdot \frac {1}{x ^ {2}} = x ^ {2} + \frac {c}{x ^ {2}}$$ $$y (1) = 1 + c = 2 \quad \Rightarrow \quad c = 1$$

Answer: $y = x^{2} + \frac{1}{x^{2}}$ , $x > 0$

Question

9. Solve for $\frac{dy}{dx} + y = 5\sin 2x$

$$y = ce2x - \sin 2x - 2\cos 3x$$ $$y = ce - x + \tan 2x - 2\cos 2x$$ $$y = ce - x + \sin 2x + 3\cos 2x$$ $$y = ce - x + \sin 2x - 2\cos 2x$$

Solution

$$\frac {d y}{d x} + y = 5 \sin 2 x$$ $$y = u v ; \quad y ^ {\prime} = u ^ {\prime} v + u v ^ {\prime}$$ $$u v + u ^ {\prime} v + u v ^ {\prime} = 5 \sin 2 x$$ $$u'v + u(v + v') = 5\sin2x$$ $$\left\{ \begin{array}{l} v + v' = 0 \\ u'v = 5\sin2x \end{array} \right.$$ $$v + v' = 0 \Rightarrow \frac{dv}{dx} = -v \quad \Rightarrow \quad \int \frac{dv}{v} = -x \quad \Rightarrow \quad \ln v = -x \quad \Rightarrow \quad v = e^{-x}$$ $$\frac{du}{dx} e^{-x} = 5\sin2x$$ $$u = \int 5e^x \sin(2x) \, dx = e^x (\sin2x - 2\cos2x) + c$$ $$y = uv = e^{-x} \left( e^x (\sin2x - 2\cos2x) + c \right)$$

Answer: $y = ce^{-x} + \sin2x - 2\cos2x$.

Question

10. Given the differential equation $\frac{dy}{dx} + \frac{1}{x}y = 3\cos2x$, $x > 0$ the solution is

$$\begin{array}{l} y = cx + 3\cos2x4x + 3\sin22 \\ y = cx + 3\cos2x4x \\ y = cx + 3\sin22 \\ y = c2x + 3\cos2x4x + 3\sin62 \\ \end{array}$$

Solution

$$\frac{dy}{dx} + \frac{1}{x}y = 3\cos2x, \quad x > 0$$ $$y = uv; \quad y' = u'v + uv'$$ $$\frac{1}{x}uv + u'v + uv' = 3\cos2x$$ $$u'v + u\left(\frac{1}{x}v + v'\right) = 3\cos2x$$ $$\left\{ \begin{array}{l} \frac{1}{x}v + v' = 0 \\ u'v = 3\cos2x \end{array} \right.$$ $$\frac {dv}{dx} = -\frac {v}{x} \quad \Rightarrow \quad \int \frac {dv}{v} = -\int \frac {dx}{x} \quad \Rightarrow \quad \ln v = -\ln x \quad \Rightarrow \quad v = \frac {1}{x}$$ $$\frac {du}{dx} \cdot \frac {1}{x} = 3 \cos 2x$$ $$u = \int 3x \cos (2x) \, dx = \frac {3}{4} (2x \sin 2x + \cos 2x) + c$$ $$y = uv = \frac {1}{x} \left(\frac {3}{4} (2x \sin 2x + \cos 2x) + c\right)$$

Answer: $y = \frac{c}{x} + \frac{3}{2} \sin 2x + \frac{3 \cos 2x}{4x}$ .

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