Question Statement

Verify that the indicated function is a solution of the given differential equation:

(i) $2y^{\prime }+y=0;y=e^{-\frac{x}{2}}$

(ii) $\frac{dy}{dx}-2y=e^{3x};y=e^{3x}+10e^{2x}$

(iii) $y^{\prime }=25+y^2;y=5\tan 5x$

(iv) $y^{\prime }+y=\sin x;y=\frac{1}{2}\sin x-\frac{1}{2}\cos x+10e^{-x}$

(v) $x^{3}dy-2dx=0;y=\frac{-1}{x^2}+6$

(vi) $y^{\prime }-\frac{1}{x}y=1;y=x\ln x,x>0$

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Background and Explanation

To verify if a function is a solution to a differential equation, we find the necessary derivatives of the given function and substitute both the function and its derivatives into the differential equation. If the substitution results in an identity (where the left-hand side equals the right-hand side), the function is a valid solution.

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Solution

Part (i)

Given Equation: $2y^{\prime }+y=0$ (Eq. 1)
Proposed Solution: $y=e^{-\frac{x}{2}}$

First, we differentiate $y$ with respect to $x$ using the chain rule: $\begin{array}{rl}{y}^{\prime }& =e^{-\frac{x}{2}}\cdot \frac{d}{dx}\left(\frac{-x}{2}\right)\\ & =e^{-\frac{x}{2}}\cdot \left(\frac{-1}{2}\right)\\ y^{\prime }& =\frac{-1}{2}{e}^{-\frac{x}{2}}\end{array}$

Now, substitute the values of $y$ and $y^{\prime }$ into Eq. 1: $\begin{array}{rl}2\left(\frac{-1}{2}{e}^{-\frac{x}{2}}\right)+e^{-\frac{x}{2}}& =0\\ -e^{-\frac{x}{2}}+e^{-\frac{x}{2}}& =0\\ 0& =0\end{array}$ The equation is satisfied. Hence, $y=e^{-\frac{x}{2}}$ is a solution.

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Part (ii)

Given Equation: $\frac{dy}{dx}-2y=e^{3x}$ (Eq. 1)
Proposed Solution: $y=e^{3x}+10e^{2x}$

Differentiate $y$ with respect to $x$: $\begin{array}{rl}\frac{dy}{dx}& =e^{3x}\cdot 3+10e^{2x}\cdot 2\\ \frac{dy}{dx}& =3e^{3x}+20e^{2x}\end{array}$

Substitute $y$ and $\frac{dy}{dx}$ into Eq. 1: $\begin{array}{rl}(3e^{3x}+20e^{2x})-2(e^{3x}+10e^{2x})& =e^{3x}\\ 3e^{3x}+20e^{2x}-2e^{3x}-20e^{2x}& =e^{3x}\\ e^{3x}& =e^{3x}\\ \text{L.H.S}& =\text{R.H.S}\end{array}$ The equation is satisfied. Hence, $y=e^{3x}+10e^{2x}$ is a solution.

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Part (iii)

Given Equation: $y^{\prime }=25+y^2$ (Eq. 1)
Proposed Solution: $y=5\tan 5x$

Differentiate $y$ with respect to $x$: $\begin{array}{rl}{y}^{\prime }& =5{\sec }^{2}5x\cdot \frac{d}{dx}(5x)\\ y^{\prime }& =5{\sec }^{2}5x\cdot (5)\\ y^{\prime }& =25{\sec }^{2}5x\end{array}$

Substitute $y$ and $y^{\prime }$ into Eq. 1: $\begin{array}{rl}25{\sec }^{2}5x& =25+(5\tan 5x{)}^2\\ 25{\sec }^{2}5x& =25+25{\tan }^{2}5x\\ 25{\sec }^{2}5x& =25(1+{\tan }^{2}5x)\end{array}$ Using the trigonometric identity $1+{\tan }^2\theta ={\sec }^2\theta$: $\begin{array}{rl}25{\sec }^{2}5x& =25{\sec }^{2}5x\\ \text{L.H.S}& =\text{R.H.S}\end{array}$ The equation is satisfied. Hence, $y=5\tan 5x$ is a solution.

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Part (iv)

Given Equation: $y^{\prime }+y=\sin x$ (Eq. 1)
Proposed Solution: $y=\frac{1}{2}\sin x-\frac{1}{2}\cos x+10e^{-x}$

Differentiate $y$ with respect to $x$: $\begin{array}{rl}{y}^{\prime }& =\frac{1}{2}\cos x-\frac{1}{2}(-\sin x)+10e^{-x}\cdot (-1)\\ y^{\prime }& =\frac{1}{2}\cos x+\frac{1}{2}\sin x-10e^{-x}\end{array}$

Substitute $y$ and $y^{\prime }$ into Eq. 1: $\begin{array}{rl}(\frac{1}{2}\cos x+\frac{1}{2}\sin x-10e^{-x})+(\frac{1}{2}\sin x-\frac{1}{2}\cos x+10e^{-x})& =\sin x\\ \left(\frac{1}{2}+\frac{1}{2}\right)\sin x+\left(\frac{1}{2}-\frac{1}{2}\right)\cos x+(-10+10)e^{-x}& =\sin x\\ \sin x& =\sin x\\ \text{L.H.S}& =\text{R.H.S}\end{array}$ The equation is satisfied. Hence, $y=\frac{1}{2}\sin x-\frac{1}{2}\cos x+10e^{-x}$ is a solution.

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Part (v)

Given Equation: $x^{3}dy-2dx=0⟹x^3\frac{dy}{dx}=2$ (Eq. 1)
Proposed Solution: $y=\frac{-1}{x^2}+6=-x^{-2}+6$

Differentiate $y$ with respect to $x$: $\begin{array}{rl}\frac{dy}{dx}& =-(-2x^{-3})+0\\ \frac{dy}{dx}& =2x^{-3}\\ \frac{dy}{dx}& =\frac{2}{x^3}\end{array}$

Substitute $\frac{dy}{dx}$ into Eq. 1: $\begin{array}{rl}{x}^3\left(\frac{2}{x^3}\right)& =2\\ 2& =2\end{array}$ The equation is satisfied. Hence, $y=\frac{-1}{x^2}+6$ is a solution.

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Part (vi)

Given Equation: $y^{\prime }-\frac{1}{x}y=1$ (Eq. 1)
Proposed Solution: $y=x\ln x,x>0$

Differentiate $y$ with respect to $x$ using the product rule: $\begin{array}{rl}{y}^{\prime }& =\frac{d}{dx}(x)\cdot \ln x+x\cdot \frac{d}{dx}(\ln x)\\ y^{\prime }& =1\cdot \ln x+x\cdot \frac{1}{x}\\ y^{\prime }& =\ln x+1\end{array}$

Substitute $y$ and $y^{\prime }$ into Eq. 1: $\begin{array}{rl}(\ln x+1)-\frac{1}{x}(x\ln x)& =1\\ \ln x+1-\ln x& =1\\ 1& =1\end{array}$ The equation is satisfied. Hence, $y=x\ln x$ is a solution.

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Product Rule: $\frac{d}{dx}[uv]=u\frac{dv}{dx}+v\frac{du}{dx}$
• Power Rule: $\frac{d}{dx}[x^n]=n{x}^{n-1}$
• Trigonometric Derivatives: $\frac{d}{dx}(\tan x)={\sec }^{2}x$, $\frac{d}{dx}(\sin x)=\cos x$, $\frac{d}{dx}(\cos x)=-\sin x$
• Exponential/Log Derivatives: $\frac{d}{dx}(e^{ax})=a{e}^{ax}$, $\frac{d}{dx}(\ln x)=\frac{1}{x}$
• Trigonometric Identity: $1+{\tan }^2\theta ={\sec }^2\theta$

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Summary of Steps

1. Differentiate: Find the first derivative of the proposed solution $y$ with respect to $x$.
2. Substitute: Plug the expression for $y$ and its derivative $y^{\prime }$ into the given differential equation.
3. Simplify: Perform algebraic simplification and apply trigonometric identities where necessary.
4. Verify: Check if the resulting statement is an identity (e.g., $0=0$ or $1=1$). If it is, the function is a solution.