# Find the general solution of the given differential equations 4y′′-4y′-3y=0

Find the general solution of the given differential equations
$4y\prime \prime -4y\prime -3y=0$
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Kayla Kline
Step 1
Given differential equation is
$4y{}^{″}-4{y}^{\prime }-3y=0$
Step 2
Auxiliary equation of differential equation is
$4{m}^{2}-4m-3=0$
$4{m}^{2}+2m-6m-3=0$
$⇒2m\left(2m+1\right)-3\left(2m+1\right)=0$
$⇒\left(2m+1\right)\left(2m-3\right)=0$
$⇒m=\frac{-1}{2},\frac{3}{2}$
Hence general solution of differential equation is
$y={C}_{1}{e}^{\frac{3}{2}t}+{C}_{2}{e}^{\frac{-1}{2}t}$
Step 3
General solution of given differential equation is $y={C}_{1}{e}^{\frac{3}{2}t}+{C}_{2}{e}^{\frac{-1}{2}t}$

usumbiix
Solving the characteristic equation and applying the initial conditions, the solution of the IVP is: $y\left(t\right)=-1.75{e}^{-0.5t}+2.75{e}^{1.5t}$
The IVP is given by:
$4y4{y}^{\prime }-3y=0$
The characteristic equation is:
$4{r}^{2}-4r-3=0$
Which is a quadratic equation with coefficients $a=4,b=-4,c=-3$, and thus:
$\mathrm{△}={b}^{2}-4ac={\left(-4\right)}^{2}-4\left(4\right)\left(-3\right)=64$
${x}_{1}=\frac{-b-\sqrt{\mathrm{△}}}{2a}=\frac{4-8}{8}=-0.5$
${x}_{2}=\frac{-b-\sqrt{\mathrm{△}}}{2a}=\frac{4+8}{8}=1.5$
Thus, the solution is in the format:
$y\left(t\right)=a{e}^{-0.5t}+b{e}^{1.5t}$
Applying the initial conditions:
$y\left(0\right)=1\to a+b=1$
${y}^{\prime }\left(0\right)=5\to -0.5a+1.5b=5$
From the first equation, $a=1-b$, and replacing in the second:
$-0.5a+1.5b=5$
$-0.5+0.5b+1.5b=5$
$2b=5.5$
$b=\frac{5.5}{2}$
$b=2.75$
$a=1-b=1-2.75=-1.75$
Then, the solution to the IVP is:
$y\left(t\right)=-1.75{e}^{-0.5t}+2.75{e}^{1.5t}$

karton

Step-by-step explanation:
Given:
$4{y}^{″}-4{y}^{\prime }-3y=0,y\left(0\right)=1,{y}^{\prime }\left(0\right)=5$
Now, the auxiliary equation will be given as:
$4{m}^{2}-4m-3=0$
or $4{m}^{2}+2m-6m-3=0$
or $2m\left(2+1\right)-3\left(2m+1\right)=0$
or $\left(2m-3\right)×\left(2m+1\right)=0$
therefore, $m=\frac{3}{2}$ and $m=\frac{-1}{2}$
thus, the general equation comes as: $y={C}_{1}{e}^{mx}+{C}_{2}{e}^{mx}$
or $y={C}_{1}{e}^{\frac{-1}{2}x}+{C}_{2}{e}^{\frac{3}{2}x}$
Now, at $y\left(0\right)=1$
Therefore, $1={C}_{1}{e}^{\frac{-1}{2}×0}+{C}_{2}{e}^{\frac{3}{2}×0}$
or ${C}_{1}+{C}_{2}=1\left(1\right)$
and, ${y}^{\prime }=\frac{-1}{2}{C}_{1}{e}^{\frac{-1}{2}x}+\frac{3}{2}{C}_{2}{e}^{\frac{3}{2}x}$
Also, ${y}^{\prime }\left(0\right)=5$
Thus, $5=\frac{-1}{2}{C}_{1}{e}^{\frac{-1}{2}×0}+\frac{3}{2}{C}_{2}{e}^{\frac{3}{2}×0}$
or $3{C}_{2}-{C}_{1}=10\left(2\right)$
on adding 1 and 2, we get ${C}_{1}+{C}_{2}=1$
$+\left(-{C}_{1}+3{C}_{2}\right)=10$
$4{C}_{2}=11$
or ${C}_{2}=\frac{11}{4}$
Thus, ${C}_{1}+{C}_{2}=1$
or ${C}_{1}+\frac{11}{4}=1$
or ${C}_{1}=\frac{-7}{4}$
Hence, the solution is $y={C}_{1}{e}^{mx}+{C}_{2}{e}^{mx}$
on substituting the values,
$y=\frac{-7}{4}{e}^{\frac{-1}{2}x}+\frac{11}{4}{e}^{\frac{3}{2}x}$