Question

3 Differential equations really don't know how to solve them help plzz

1) dz/dt+ e^(t + z) = 0
2) y' tan x = 7a + y, y(π/3) = 7a, 0 < x < π/2
3) xy'' + 2y' = 16x2 use the substitution u = y'

Answer 1

\[\frac{dz}{dt} + e^{t}e^{z} = 0\]
\[\frac{dz}{dt} = -e^{t}e^{z} \]
\[\frac{dz}{e^{z}} = -e^{t}dt\]
integrate both sides
\[-\frac{1}{e^{z}} = -e^{t}+C\]
\[e^{z} = \frac{1}{e^{t}+C}\]
\[z = -\ln (e^{t}+C)\]

Answer 2

ohh euh thats what i tried but it seems like its not right i dunno why..thanks though

Answer 3

\[\frac{dy}{dx}\tan(x) = 7a +y\]
\[\frac{dy}{7a+y} = \frac{dx}{\tan(x)}\]
integrate both sides
\[\ln (7a+y) = \ln (\sin x)+C\]
\[7a+y = C \sin(x)\]
\[y = C \sin(x) -7a\]
\[y(\pi/3) = 7a \rightarrow \frac{\sqrt{3}}{2}C -7a = 7a \rightarrow C = \frac{28\sqrt{3}a}{3}\]
\[y = \frac{28\sqrt{3}a}{3} \sin(x) -7a\]

Answer 4

#1 is correct
http://www.wolframalpha.com/input/?i=dz%2Fdt+%2B+e^t*e^z+%3D+0

Answer 5

Letting y' = u, y'' = du/dx

\[x \frac{du}{dx} +2u = 16x^{2}\]
divide equation by x
\[\frac{du}{dx} + \frac{2}{x} u=16x\]
This is non-separable so we need to find integrating factor
\[\large \rightarrow e^{\int\limits_{}^{}2/x} = e^{2 \ln x} = x^{2}\]
multiply equation by x^2
\[ \frac{du}{dx} x^{2} + 2x u=16x^{3}\]
\[(x^{2} u)' = 16x^{3}\]
integrate both sides
\[x^{2} u = 4x^{4} +C\]
\[u = 4x^{2}+\frac{C}{x^{2}}\]
u = y' = dy/dx
\[y = \int\limits_{}^{} 4x^{2}+\frac{C}{x^{2}} dx\]
\[y = \frac{4}{3}x^{3} +\frac{C}{x}+K\]